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00:10:00Paolo Guiotto: Okay, let me quickly, review, the definition of logarithm.
00:16:630Paolo Guiotto: Yesterday, We introduced, we introduced the logarithm of the complex logarithm, or the C-Logger.
00:29:200Paolo Guiotto: So we started from the equation E to W.
00:34:280Paolo Guiotto: equal, Z.
00:37:320Paolo Guiotto: where Z is given, and it is convenient to write in a trigonomatic form. So, raw…
00:46:450Paolo Guiotto: times cos theta plus i sine theta.
00:50:950Paolo Guiotto: That we reminded that because of the Euler formula, this is also E to i theta, so…
00:57:660Paolo Guiotto: The notation, it could be a little bit shortened into this.
01:03:790Paolo Guiotto: Well, raw is positive.
01:07:00Paolo Guiotto: or 0, and it represents the distance of the point Z to the origin on the complex plane.
01:15:640Paolo Guiotto: And theta is the angle made by the number, the Z, the vector Z, and the positive direction of the Cartesian axis. So normally, this is an angle that we
01:32:720Paolo Guiotto: For the formula, theta could be any real number, but…
01:36:60Paolo Guiotto: Normally, we consider as a number between 0 and 2 pi. So, we also… we use these notations that raw is the modulus of Z, and this particular theta, the angle between 0 and 2 pi, is called the argument
01:54:550Paolo Guiotto: of the number Z, okay? So, to give a name this in terms of Z.
02:01:80Paolo Guiotto: Now, what happens? It happens that E to Z, E to W is equal to Z, with W equal X plus IY, if and only if…
02:14:390Paolo Guiotto: First, we need the data raw be positive, so Model Z is positive.
02:21:130Paolo Guiotto: So, in other words, Z is different from 0.
02:25:270Paolo Guiotto: And for this,
02:28:130Paolo Guiotto: for this Z, we have the values for W are these ones, logarithm of modules of Z plus i theta, so the argument of Z.
02:42:650Paolo Guiotto: Plus multiple of 2 pi. Multiples, because there are infinitely many.
02:49:10Paolo Guiotto: of this, with this coefficient Karen in Z. So, we use this to call this the, logarithms. These numbers are called,
03:03:660Paolo Guiotto: Logarithms. C logarithms.
03:07:500Paolo Guiotto: off.
03:08:550Paolo Guiotto: the complex number Z, okay?
03:14:420Paolo Guiotto: So…
03:15:930Paolo Guiotto: In other words, log of Z, we may write, is equal to the log of modulus of Z, so this is the C logarithm, this is the real logarithm, plus I,
03:29:80Paolo Guiotto: argument of Zebda, so the… Angle between 0 and 2 pi, plus multiples, of Tubai.
03:39:160Paolo Guiotto: A in Z. This is the formula. So differently, we have many differences with respect to the real logarithm. For example, this logarithm is defined for every complex number, provided it is non-zero, so for every Z different from zero.
03:56:970Paolo Guiotto: And for real logarithm, this is defined only for positive numbers, okay?
04:02:840Paolo Guiotto: Second, differently from the real logarithm, where the real logarithm is a unique value, here, the logarith is infinitely many values, so there are infinitely many logarithms. At some point, we will decide to call logarithm one of them, that one with K equals 0, basically.
04:22:250Paolo Guiotto: But we have to know that the solutions of the equation E to W equals Z are these numbers. There are infinitely many solutions.
04:30:100Paolo Guiotto: And this is quite important to solve a number of equations. We can,
04:37:780Paolo Guiotto: We can solve it with elementary functions. So, let me show this. We have done an example yesterday, but… but let's see more examples, and also I left you to do something in the exercises. So, let's start with the example.
04:55:600Paolo Guiotto: 728 in notes. So, there are a few simple equations, like solve, for example, E to Z equal 1.
05:08:420Paolo Guiotto: Now, in R, you would say that this equation has solution, in R.
05:15:470Paolo Guiotto: in reels.
05:17:180Paolo Guiotto: Solution is, is zero, okay?
05:20:120Paolo Guiotto: So, remind that in R, E to X equal 1 if and only if x is equal to 0.
05:29:110Paolo Guiotto: Let's see what happens in C.
05:31:710Paolo Guiotto: Well, we have to expect that this is, E to Z equal 1 means that Z is the logarithm of 1, so, E to Z is equal to 1 if and only if Z is the C logarithm of 1.
05:48:850Paolo Guiotto: And so, since there are infinitely many C logarms, as soon as the number is different from 0, 1 is different from 0. So now, to write these numbers, we have to put 1 in trigonometric form. So 1 is the modulus of 1 is 1 to 1 is here.
06:07:380Paolo Guiotto: is on the real axis, is 1 times e to i, the argument, so the angle, which is in this case, 0. So this is the modulus of 1, and this is the argument of 1.
06:24:80Paolo Guiotto: So the formula says that we have log, real log of modulus of 1, plus I, the argument, which is 0, plus multiple of 2 pi.
06:37:430Paolo Guiotto: So at the end, since a log of modulus 1, this is 1, this logarithm is 0. So we have IK2 pi, or I, sorry.
06:48:730Paolo Guiotto: It's the same, of course, but that was to emphasize that they are product multiples of 2 pi. K in Z. So let's see in the figure where they are, these logarithms.
06:59:790Paolo Guiotto: So, 4K equals 0, This is the complex plane.
07:06:510Paolo Guiotto: For k equals 0, you see that we get that I0 is 0, so we get this one, which is the real, solution, no? But when k is different from 0, we have something like i to pi here.
07:22:120Paolo Guiotto: So this is 0, i2 pi, I4 pi.
07:26:30Paolo Guiotto: I6 pi, and so on, but also negative, coefficient negative K, so, minus i to pi
07:35:480Paolo Guiotto: minus i4 pi, and so on. So these are the logarithms. As you can see, only one of them is real, but there are infinitely many others which are pure imaginary numbers.
07:49:130Paolo Guiotto: So these are the solutions.
07:51:220Paolo Guiotto: For example, yeah?
07:54:40Paolo Guiotto: No. Mmm…
07:58:380Paolo Guiotto: No, no, because in general, they are complex. So, for example, the number 2 is, we can say that there can be at most, one real, if you want.
08:10:670Paolo Guiotto: But not always. Okay, let's look at this. C to Z equal Y, so this means that Z is the C log
08:20:230Paolo Guiotto: of number i. Now, the formula says that this is the real log of the modulus of I plus I, the argument of I,
08:33:710Paolo Guiotto: Plus multiple after pi.
08:37:30Paolo Guiotto: Now, modulus of I is 1, i is here.
08:42:940Paolo Guiotto: So modulus of i is equal to 1, and the argument of I
08:49:470Paolo Guiotto: is this angle, which is pi half.
08:54:430Paolo Guiotto: So we have that this is log of… real log of 1, plus I pi half…
09:03:330Paolo Guiotto: Plus multiple of 2 pi.
09:06:980Paolo Guiotto: So, log of 1 is 0, the real log of 1 is 0, and this is i pi half plus multiple of 2 pi K in Z.
09:20:720Paolo Guiotto: So also here we can have an idea where these numbers are in the complex plane, C.
09:28:300Paolo Guiotto: So, for k equals 0, I have i pi half, it's on the imaginary axis, about here.
09:35:310Paolo Guiotto: Then we have i pi half plus 2 pi, so we add 2 pi, let's say we are here, pi half plus 2 pi.
09:45:530Paolo Guiotto: And, and so on. We continue to add the 2 pi, and here we subtract 2 pi, so we'll be more or less here.
09:54:80Paolo Guiotto: So this is i pi half minus 2 pi.
09:59:850Paolo Guiotto: So if you want to write it in a different way, it is minus 3 half pi, so minus i3 half pi, and so on, no?
10:12:320Paolo Guiotto: So these are the solutions. As you can see here, there is no real solution, for example.
10:20:330Paolo Guiotto: Let's see something a little bit more spicy. It's not… well, let's see…
10:31:450Paolo Guiotto: I want to go to the exercises.
10:43:330Paolo Guiotto: No, it is not this one, so… This is another example.
10:49:670Paolo Guiotto: Suppose we want to solve something that in real is impossible, so hyperbolic cosine of Z is equal to 0. I remind you that in R,
11:01:140Paolo Guiotto: Hyperbolic cosine of X equals 0 is never.
11:05:930Paolo Guiotto: It's impossible.
11:08:860Paolo Guiotto: The hyperbolic cosine is always greater or equal than 1, when the variable is real. But, as we will see, hyperbolic cosine of 0 can be possible when we are in Z.
11:21:40Paolo Guiotto: How do we solve? Basically, as you can see, everything depends on the exponential, and the exponential yields sooner or later to a logarithm. So, that's the strategy, no? So, we replace
11:33:960Paolo Guiotto: the hyperbolic cosine of Z by its definition, which is e to Z plus e to minus Z divided by 2. So we want that this be equal to 0.
11:45:20Paolo Guiotto: Then this means that we divide, we multiply by 2. E to Z plus E to minus Z is 1 over e to Z equals 0. We remind that also for the exponential, for the complex exponential, it is never equal 0, okay?
12:01:880Paolo Guiotto: Have you tried to prove that?
12:08:130Paolo Guiotto: In order to be equal to zero, it should be zero, but we have the E that is equal to the exponential, and at the moment is zero, which is equal to 0. No, because that's just the notation.
12:20:20Paolo Guiotto: So, it's a little bit more subtle. E to Z is always different from 0 for every Z, but it's just one step, because if you multiply E to Z times e to minus Z, the two are well defined, you know, are two exponential. What do we get? Because of the properties of the exponential, we get that this is E to Z minus Z, right?
12:42:690Paolo Guiotto: So, e to 0.
12:44:250Paolo Guiotto: E to 0 is 1, as we have seen, so the product of these two guys is 1, so it means that none of them can be 0. So in particular, e to Z and E2 minus Z are both different from 0. That's it.
12:58:920Paolo Guiotto: Okay?
13:00:180Paolo Guiotto: So, the exponential, even if Z… you see, it's not evident, because E to Z formally, is the sum of this series, Z to the n divided then factorial for n going from 0 to infinity. How can you solve this equation? It's… it's not easy to solve this equation directly from that formula.
13:22:260Paolo Guiotto: No, nobody can solve, but actually, we can… we can see that it is never verified because of this simple argument, okay?
13:33:560Paolo Guiotto: Okay, since it is always different from 0, 1 over E to Z is well-defined. We multiply both sides by e tox, E to Z, sorry, which is different from 0, so we do not change the equation, and this becomes E to 2Z plus 1 equals 0, which is now a logarithmic equation, because e to 2Z
13:56:680Paolo Guiotto: plus 1 equals 0, if and only if e to 2Z is equal to minus 1.
14:03:890Paolo Guiotto: So it means that 2Z is the C logarithm of minus 1.
14:12:320Paolo Guiotto: And this is the real logarithm of modules of minus 1.
14:18:670Paolo Guiotto: plus I, the argument of minus 1, plus multiples of 2 pi.
14:27:310Paolo Guiotto: Now, minus 1 is in the real axis, but on the negative part.
14:33:380Paolo Guiotto: So the modules of minus 1 is 1, and the log here is 0, and the argument is this angle, which is 180 degrees, or pi in gradients. So we get that this is I,
14:47:880Paolo Guiotto: pi plus 2, k to pi, or if you want, It is I, pi.
15:01:720Paolo Guiotto: times an odd number, 2K plus 1 with K in Z.
15:08:810Paolo Guiotto: So, these are what… these are the solutions of e to 2Z equals minus 1, which is, at the end, equivalent to cosh equals 0. So, we can say that cosh…
15:20:360Paolo Guiotto: Z equals 0 if and only if Z is equal I, 2K plus 1,
15:29:540Paolo Guiotto: pi with K in Z, and these are the solutions of this equation. Huh?
15:40:740Paolo Guiotto: 2Z, you're right. So this will be divided by… Truth.
15:46:420Paolo Guiotto: Maybe, but it's fine.
15:49:280Paolo Guiotto: Thank you.
15:52:570Paolo Guiotto: So these are the solutions of this equation.
15:57:360Paolo Guiotto: So now we can… we can also draw, in this case, where are these points in the C plane. So for k equals 0, we get I pi half, so it's about here.
16:09:270Paolo Guiotto: then we have to do pi half, maybe it's better if we write in this way. It is I…
16:16:200Paolo Guiotto: Mmm… but, let's say… mmm… by half.
16:23:610Paolo Guiotto: plus, a multiple of pi. So we have to add… pi, k times,
16:32:580Paolo Guiotto: Or subtracted. So the next one will be, say, here, pi half plus pi is 3 halves.
16:39:460Paolo Guiotto: pi, then we have I5 half phi, and so on. Here we have minus i pi half…
16:49:10Paolo Guiotto: minus, I3 half pi, and so on.
16:54:240Paolo Guiotto: So these are the solutions of this equation. As you can see, none of this solution is real, because on R, there are no solutions for this equation. But in C, there are.
17:10:160Paolo Guiotto: Okay, so let's see if you tried to do some of the exercises at 7… 11… True.
17:21:730Paolo Guiotto: For example, here, these are equations with unknown Z, so the number one says cosh square Z
17:32:250Paolo Guiotto: plus 1 equals 0. You see that if we are in R, so Z is a real number, this equation has no solutions, because cosh squared is positive, plus 1 cannot be 0. But we are in C, and in C, the situation is different.
17:48:700Paolo Guiotto: Because this means that cosh square z is equal to minus 1. So, it means that cosh square
17:57:620Paolo Guiotto: Sorry, Kosh Zeda…
18:00:510Paolo Guiotto: is one of the root of minus 1, and there are two roots which are plus
18:07:90Paolo Guiotto: So, plus minus the root of minus 1, which are plus minus i.
18:11:580Paolo Guiotto: So basically, we have to look for cosh z equals i, or minus i. So let's solve the first one. Cosh Z equals I means… well, now you plug the definition of cosh, which is e to z plus e to minus Z divided 2.
18:30:50Paolo Guiotto: is equal to I
18:32:10Paolo Guiotto: And this now is an equation in the quantity e to Z that we can easily solve. We multiply by 2.
18:40:200Paolo Guiotto: Then we rewrite as E to Z plus 1 over e to Z, same argument as above. This is different from 0. We multiply everything by e to Z.
18:53:420Paolo Guiotto: which is different from 0. This is important, because when we divide, we multiply an equation by something. We have to be sure that it is not 0, otherwise we change the story of the equation. So this is now equivalent to e to 2Z plus 1
19:09:210Paolo Guiotto: equal I to E to Z. So it means that the quantity E to Z fulfills this equation, E to Z squared minus i to e to Z
19:21:740Paolo Guiotto: plus 1 equals 0. And this is a second degree equation in the quantity e to Z, no? W squared minus i2W plus 1 equals 0.
19:34:220Paolo Guiotto: Fortunately, algebraic equations have always solutions in C, differently from the reals, and the solutions in this case are given by the usual formula. There are two solutions, W12 equal, so minus that, so I2 plus minus the root of
19:53:210Paolo Guiotto: That's right, slowly minus…
19:55:310Paolo Guiotto: 2i squared minus 4 times 1 times 1. This is the argument of the root, divided to the coefficient of W squared, which is 1.
20:06:100Paolo Guiotto: So… This yields, 2I plus minus.
20:12:290Paolo Guiotto: Root. Off.
20:14:390Paolo Guiotto: Minus 1 squared is plus 1. I squared is minus, so minus…
20:19:950Paolo Guiotto: 2 squared is 4, minus another 4, so divided 2, so this is minus 8.
20:27:550Paolo Guiotto: So we can say that the root of minus 1 is i, so we have I2 plus minus i root of 8,
20:35:480Paolo Guiotto: divided 2. This can be seen as 4 times 2, so 2 root of 2, that yields slightly to simplify this quantity into I,
20:48:640Paolo Guiotto: to… ugh.
20:51:360Paolo Guiotto: 1… Plus minus root of 2.
20:55:860Paolo Guiotto: And this is the W.
20:57:960Paolo Guiotto: Okay, two values of W, they are E to Z. So we have to solve E to Z equal this quantity.
21:05:680Paolo Guiotto: Now, this means that Z is the C logarithm
21:11:770Paolo Guiotto: of this complex number, i1 plus minus root of 2.
21:20:370Paolo Guiotto: Okay?
21:22:410Paolo Guiotto: So, these two numbers, i plus minus… I times 1 plus minus root of 2. Clearly, the one with plus is this one, 1 plus root of 2 is on the positive
21:39:360Paolo Guiotto: Part of the imaginary axiom.
21:41:900Paolo Guiotto: The first… the second one with the minus, root of 2 is greater than 1, so 1 minus root of 2 is negative, so it's about here.
21:50:780Paolo Guiotto: So, I1 minus the root of 2. This is important because to compute the logarithm, I need a trigonometric representation of these numbers. So, what is the trigonometric representation of I1 plus root of 2? It is the modulus of this number. The modulus is
22:15:670Paolo Guiotto: No, the modulus. Be careful, because argument means an angle here, okay? The modulus is…
22:23:870Paolo Guiotto: Yeah, so it's the 1 plus root of 2.
22:28:20Paolo Guiotto: And the argument is… Bye.
22:33:900Paolo Guiotto: For the second one, I1 minus root of 2, the modulus is
22:42:880Paolo Guiotto: My husband blessed me.
22:45:340Paolo Guiotto: So, it's root of 2 minus 1.
22:48:580Paolo Guiotto: And the argument is… No, because it is this one.
22:54:300Paolo Guiotto: Is 3… 3 afts.
22:58:480Paolo Guiotto: Okay, so this means that if I have to compute the C logarithm of that one with plus, I1 plus root of 2,
23:11:970Paolo Guiotto: This is the real logarithm.
23:15:380Paolo Guiotto: of 1 plus… root of 2, plus I, the argument, which is pi half, plus multiple of 2 pi.
23:27:400Paolo Guiotto: We cannot simplify any…
23:29:910Paolo Guiotto: Better than this. And for the second one, we have log C of i, 1 minus root of 2,
23:38:670Paolo Guiotto: This is the real logarithm of… now, it is the root of 2 minus 1,
23:44:490Paolo Guiotto: plus I, the argument is 3 half pi, plus multiples of 2 pi.
23:52:640Paolo Guiotto: A EZ.
23:55:150Paolo Guiotto: Okay, now, what are these?
23:58:380Paolo Guiotto: These are the solutions to this equation.
24:02:820Paolo Guiotto: Which are the solutions to this equation.
24:08:460Paolo Guiotto: which are the solution to cosh Z equals I.
24:14:720Paolo Guiotto: So, are not yet the total solutions, because we have to solve also for cosh equal minus i, okay? So, we can say, let's summarize here, cosh z equal i if and only if
24:29:510Paolo Guiotto: Z is equal, either one of the two. Log…
24:34:930Paolo Guiotto: Of 1 plus root of 2.
24:39:90Paolo Guiotto: Last I, pi a half.
24:42:150Paolo Guiotto: plus K 2 pi, this for K.
24:46:60Paolo Guiotto: inside the… or… Z also log, real log, of root of 2.
24:54:220Paolo Guiotto: minus 1 plus I, 3 halves pi, plus K2 pi.
25:01:900Paolo Guiotto: This 4K is at.
25:03:670Paolo Guiotto: Now we have to solve the second equation, caution.
25:08:280Paolo Guiotto: Zed… Equal, minus… I… Well, as you may imagine, it's not particularly
25:20:360Paolo Guiotto: is the same of the previous equation, so what we do is we write e to z plus e to minus Z divided by 2 equal to minus i. This is equivalent. We do some algebra, we multiply by 2,
25:36:520Paolo Guiotto: We multiply also directly by E2 minus Z, t2 plus Z, so we get e to Z squared.
25:45:590Paolo Guiotto: plus 1 equal minus i to E to Z.
25:52:10Paolo Guiotto: And this boils down to the equation E to Z squared.
25:57:510Paolo Guiotto: plus, now, I2E to Z.
26:01:510Paolo Guiotto: loss.
26:02:640Paolo Guiotto: 1 equals 0.
26:05:120Paolo Guiotto: So, we have to solve the second degree equation W squared plus i to W plus 1 equals 0, which is more or less the same equation, so let's see if we can
26:18:550Paolo Guiotto: Use the same calculations.
26:24:580Paolo Guiotto: Yeah, you see the… there? Well, let's write the formula.
26:28:360Paolo Guiotto: So W12 is equal to 2. Minus I2 plus minus the root of I2 square, which is still minus 4, and then we have minus 4, minus 4 times 1, 1.
26:42:620Paolo Guiotto: divided by 2, so it's minus i2 plus minus, again, 2i, 2 root of 2 divided by 2.
26:51:720Paolo Guiotto: So, simplifying the two… and factorizing a minus, we have minus i times 1 plus minus… root of 2.
27:05:620Paolo Guiotto: Right?
27:08:580Paolo Guiotto: Yeah, they are not exactly the same, because now these two points are where… when I take the plus, it is here.
27:17:410Paolo Guiotto: This is minus i.
27:20:160Paolo Guiotto: 1 plus root of 2.
27:24:530Paolo Guiotto: And the second one…
27:26:680Paolo Guiotto: it is minus i times 1 minus root of 2, but that's negative, so with the minus becomes positive, so it is here. So if you want, we can write i times root of 2 minus 1.
27:41:160Paolo Guiotto: So they are, like that.
27:43:970Paolo Guiotto: So, these W are the values of E to Z, so we have to solve. E to Z equal i root of 2 minus 1,
27:53:920Paolo Guiotto: That means this is Z equal log of… real log of the modulus of this number, which is the root of 2 minus 1.
28:06:250Paolo Guiotto: plus I, the argument of this is pi half, pi half plus multiple of 2 pi.
28:14:940Paolo Guiotto: And the second one is the log R. The modus is now still 1 plus root of 2.
28:22:470Paolo Guiotto: Because this distance is 1 plus root of 2, plus I, the argument for this one is 3 half pi.
28:30:610Paolo Guiotto: So I3 half pi plus multiple of 2 pi.
28:36:340Paolo Guiotto: A Z.
28:40:160Paolo Guiotto: They look the same of the previous number, but they are not the same, I suspect.
28:46:950Paolo Guiotto: Because, here we have log 1 plus root of 2, so the real part is log of 1 plus root of 2, that would be this one. But then we have I3 half pi plus multiple of two pi. Maybe we can unify the two, that's just…
29:04:750Paolo Guiotto: Because there we have a pi half plus multiples of 2 pi, no? So, pi half, then, if you look at the angles, this is pi half, then you have pi half plus 2 pi, then plus pi half plus 4 pi, and so on.
29:23:280Paolo Guiotto: But here you have 3 halves pi.
29:27:480Paolo Guiotto: Which is exactly pi half plus pi, so it's here.
29:32:820Paolo Guiotto: 3 half pi. Then we have to add 2 pi, so, this will be 3 half pi plus 2 pi.
29:41:590Paolo Guiotto: So this angle here, together with the other ones, means that we are taking pi half plus multiple of pi. So if you want, we can, find a conclusion, cosh z equal…
29:58:30Paolo Guiotto: Now, of course I do not remember what is the question. Cash Square.
30:04:290Paolo Guiotto: Tosh squared plus 1 equals 0, if and only if… so I can say z is equal to log real of 1 plus root of 2.
30:16:380Paolo Guiotto: plus I, I just write pi half plus a multiple of pi, because when I put together these ones.
30:25:400Paolo Guiotto: with that.
30:28:320Paolo Guiotto: these ones…
30:30:700Paolo Guiotto: I can write in a unified way, it's just the simplification of the writing, that's not anything substantial.
30:38:180Paolo Guiotto: And for the other, I have the same music… real logarithm of root of 2 minus 1 plus I, and it will be…
30:48:750Paolo Guiotto: Yeah, the same music, because also, yeah, we have 3 half pi plus multiples of 2 pi, and here pi half plus multiple… so the angles are still the same, pi half plus multiples of…
31:00:600Paolo Guiotto: Bye.
31:01:780Paolo Guiotto: So these are all the possible solutions of that equation, okay?
31:09:310Paolo Guiotto: It's a bit complicated, but at the end, the complication is just calculations. The logarithm you see, it's just at the end, where you have to solve E to Z equals something.
31:24:590Paolo Guiotto: So, I don't want to exaggerate to insist, but this could be, typically, an exercise you can find in an exam, so it's important that you are able to solve these equations. For example.
31:37:220Paolo Guiotto: Let's do another one. The number 5. Cosine…
31:42:670Paolo Guiotto: Z equal I, something that you don't see in reels.
31:50:70Paolo Guiotto: So now, how can we do? The same thing, because everything can be defined through the exponential. So, we know that, we know…
32:00:810Paolo Guiotto: that…
32:03:30Paolo Guiotto: cos z is equal, the formula is similar to the cosh, but with the argument which is IZ, so it is…
32:11:700Paolo Guiotto: E2IZ plus E2 minus iZ divided by 2. Be careful, because 4 sine is… we don't need it here, but 4 sine…
32:23:170Paolo Guiotto: Slightly different, because there is 2Y in the denominator.
32:28:910Paolo Guiotto: So, this is the formula.
32:32:640Paolo Guiotto: By the way, I take this opportunity to say to you that,
32:38:600Paolo Guiotto: Well, I will let you to carry a paper, say, where you can write formulas you think are important, okay? I can check what you have with you.
32:50:300Paolo Guiotto: So, please do not write solutions of exercises, even… you have only one paper to use, so you… even if you…
33:00:610Paolo Guiotto: If you write with a microscope, it would be… it would be impossible to read, at least for me. So, just formulas.
33:11:540Paolo Guiotto: be responsible, so, I will… I will check, potentially, anything you carry with you.
33:18:810Paolo Guiotto: We, addicta.
33:22:520Paolo Guiotto: I see, yes, just like this room, right? I don't know what papers are used to. Normally, A4 is the type of paper.
33:34:780Paolo Guiotto: Oh, if you want, we could use the US standard letter paper, which is slightly smaller, if you want.
33:43:390Paolo Guiotto: So, no, because when… normally, when you… when you… when you send something in to an American, guy, no,
33:53:390Paolo Guiotto: So, normally, you prepare the document in a full format.
33:58:520Paolo Guiotto: And when this person prints the document, it turns out that the printer won't printed, it just cut off the page, because they print on these different papers, so they normally do not like too much A4, they prefer
34:14:389Paolo Guiotto: These other formats.
34:15:800Paolo Guiotto: Okay, so let's say that cos Z is equal to I if, and only if EIZ plus E minus iZ divided by 2 is equal to I.
34:30:730Paolo Guiotto: So basically, we got the same equation. We did the above, but just the IZ… In place of,
34:40:70Paolo Guiotto: So, we can use, I will repeat, because it's like if we have never seen the solution we have done here, okay?
34:48:520Paolo Guiotto: So this is,
34:52:310Paolo Guiotto: This is now multiplying by E2IZ, which is different from 0, and the exponential is different from 0. We get e to iZ square.
35:04:160Paolo Guiotto: plus 1 equals i to E2IZ.
35:09:980Paolo Guiotto: So, carrying everything at the left-hand side, we have e to iZ square minus i2, e to iZ plus 1.
35:18:950Paolo Guiotto: So that's the equation, W squared minus i2W plus 1 equals 0, since we solved, but we do not save too much time, but I will spend more time in watching the solution, so I will redo the solution.
35:35:800Paolo Guiotto: So W1 to me is I2 plus minus the root of minus I2 squared, which is minus 4, minus 4 here, divided 2, so it is… so at the end, I…
35:50:490Paolo Guiotto: times 1 plus minus the root of 2. This is the solution. So we have…
35:57:410Paolo Guiotto: This is W. W is this guy, E-I-Z.
36:03:250Paolo Guiotto: must be equal to I1 plus minus the root of 2.
36:09:590Paolo Guiotto: So, the unique difference is that now we have that IZ,
36:14:760Paolo Guiotto: is the logarithm, the C logarithm of i1 plus minus root of 2.
36:23:890Paolo Guiotto: So if we want Z, we have to divide by I.
36:28:130Paolo Guiotto: Now, remind that fast shortcuts, nothing special. When you divide by i, so you multiply by 1 over i, 1 over i is equal to
36:41:910Paolo Guiotto: minus i, okay? So dividing by i is like multiplying by minus i. So we have Z is equal to minus i, the C log
36:54:110Paolo Guiotto: of these numbers, I1 plus minus root of 2.
36:59:970Paolo Guiotto: So, for example, Z equal minus i, the C log of i
37:08:80Paolo Guiotto: 1 plus root of 2. Let's do this one, the other one is the same. We have already seen what is this, no? We have, so minus i times the real log of the modulus, which is… yeah.
37:23:590Paolo Guiotto: It would be 4 top 2 plus minus 1, so we would have to apply two different answers.
37:36:780Paolo Guiotto: No, I do not understand your question.
37:39:780Paolo Guiotto: instead of writing 1 plus, the root of 2, and then another answer, the root of 2, and minus 1, just write root of 2 plus and minus 1. So, we combine the two answers.
37:56:310Paolo Guiotto: Brutal 2?
38:00:300Paolo Guiotto: Yeah, I was talking about this, no?
38:03:20Paolo Guiotto: Sure.
38:04:410Paolo Guiotto: bottle?
38:13:630Paolo Guiotto: Yeah, but you see, the point is that the argument is different, because the modulus, it's clearly modulus of 1 plus minus root of 2, okay? So…
38:24:170Paolo Guiotto: Who cares? But the argument is different, because that one with plus.
38:29:310Paolo Guiotto: We did this discussion here, they are these two points. The one with plus is in the positive imaginary imaginary axis, so the argument is pi half. The other with minus is in the negative, so the argument is entirely different. Here we have pi half, and this is 3 half, 3 pi half.
38:48:700Paolo Guiotto: I'll probably love this.
38:52:270Paolo Guiotto: Yeah, but I don't see such a particular… yeah, I can put modulus 1 plus minus, but then I have two different, objects later. You see, I cannot skip that. So, I can say this is 1 plus…
39:10:800Paolo Guiotto: However, you do as you like, as you prefer. The unique rule is that it must be correct. If you write something wrong, I will be nasty with you.
39:20:780Paolo Guiotto: Okay, so this is pi half plus multiple of 2 pi.
39:28:60Paolo Guiotto: Because at the end, you see, even here, with this previous exercise, there is another interesting curiosity that we simplified, you know, at the end, the result, you know? Instead of four different formulas for that, we
39:41:720Paolo Guiotto: we just simplified into two. But, what… what combines are different, no? So…
39:49:720Paolo Guiotto: Different cases, so it's something that…
39:53:170Paolo Guiotto: I don't know what is convenient to do, really. However, this is the formula for the case with plus. Now, we multiply by i, and here you discover that what is the imaginary part is the part that contains this real logarithm.
40:08:70Paolo Guiotto: And then, the minus i times i is plus 1. I squared is minus 1 with minus is plus 1 plus, and this is now the real part, 5 halves plus K2 pi.
40:21:330Paolo Guiotto: So if I have to see where these values are in the complex plane, the real part is pi half plus multiples of 2 pi.
40:33:820Paolo Guiotto: And there is an imaginary part. Now, log of 1 plus root of 2 is positive, the real log, because the argument is greater than 1. So this is a positive number. With the minus, it means that this is a negative imaginary part. So let's say that here we have the ordinate, which is the value minus log of 1 plus root of 2.
40:55:390Paolo Guiotto: And, now, this is the ordinate. The assistants are pi half plus… so, means that if this is pi half, for example.
41:04:340Paolo Guiotto: This is one of the solutions in here. Then we have pi half plus 2 pi.
41:11:800Paolo Guiotto: So, the point will be down here, and then we have another point here, another here, and then we have, similarly here, minus 2 pi, minus 2 pi for the abyssis. So, in this case, these are the solutions.
41:25:840Paolo Guiotto: for this part of the story. Then there is a second part with the other, with the other logarithm, with the minus.
41:33:120Paolo Guiotto: So we have also Zed, equal minus psi.
41:43:190Paolo Guiotto: log… C logarithm of what is i times 1 minus root of 2. So yes, this is minus i, the real logarithm of the modulus, you could have, but at the end.
41:59:870Paolo Guiotto: I don't see what you gain by doing this.
42:03:840Paolo Guiotto: And then plus I. Now, this time, this number is that one with the negative,
42:09:590Paolo Guiotto: Imaginary part, so the angle is 3 half pi plus multiples of 2 pi.
42:16:770Paolo Guiotto: And again, when you multiply everything by this factor minus i, you get that this is 3 half pi plus K2 pi. This is the real part.
42:29:110Paolo Guiotto: Minus i, log of root of 2 minus 1.
42:36:290Paolo Guiotto: Now, where these numbers are.
42:40:10Paolo Guiotto: You see that they have, now affixed the imaginary part.
42:45:700Paolo Guiotto: The number log root of 2 minus 1 is what? Root of 2 is 1.4, let's say, no? Minus 1 is 0.4, so log is negative, with the minus there, means that the imaginary part is positive. So the value minus log 2
43:02:410Paolo Guiotto: of root of 2 minus 1 is, let's say, about here. Then for the abscesses, we have 3 half pi, so let's say that this is 3 half pi.
43:13:820Paolo Guiotto: Okay, so the first, with k equals 0 is 3 of pi minus i log of, etc. So it is a point like here.
43:23:380Paolo Guiotto: Then I have to add 2 pi, then I have to add another 2 pi, or subtract 2 pi, so maybe here, here, here. So these are the solutions. You can see we cannot unify with that one.
43:37:750Paolo Guiotto: Okay.
43:44:660Paolo Guiotto: Okay.
43:47:530Paolo Guiotto: Let's do one last… a last one, number 6.
43:52:880Paolo Guiotto: But they are, as you can see, very similar, so the unique thing to do is to practice, so please do these exercises. E to Z squared equal 1. Now, if we were in reals, E to X squared equal 1,
44:09:150Paolo Guiotto: Exponential is 1 when the argument is 0, so Z square equals 0, Z equals 0. But in complex numbers, it's different.
44:16:550Paolo Guiotto: Because this means that,
44:19:00Paolo Guiotto: The exponential of this number is 1. That number, Z squared, is the C logarithm of 1. What is the C logarithm of 1? It's the real logarithm of 1, of the modulus of 1,
44:33:500Paolo Guiotto: plus I, the argument of 1 is
44:37:940Paolo Guiotto: 0 plus multiples of pi, 2 pi.
44:43:410Paolo Guiotto: So, all this is 0, so at the end, this points down to IK2 pi.
44:51:110Paolo Guiotto: with K in Z.
44:54:830Paolo Guiotto: Now, we have to solve Z square equal i to k to pi.
45:02:710Paolo Guiotto: with K inside.
45:05:460Paolo Guiotto: Which is something that in real, apart for k equals 0, has no meaning, but in complex has a meaning, okay?
45:13:610Paolo Guiotto: So, here, maybe we have to be a little bit careful in the discussion, because for k equals 0, we have something a slightly different and simple discussion. For k equals 0, you have Z squared equals 0, so this is clearly Z equals 0.
45:31:640Paolo Guiotto: Because the product is the product, and the product is zero, if and only if one of the factors, they are the same is 0.
45:38:990Paolo Guiotto: For K, different from 0, basically the problem is simple. I have to compute the roots, the square roots of these numbers.
45:47:550Paolo Guiotto: But since to compute the roots, so let me just remind that…
45:57:130Paolo Guiotto: In general, this is the Des Muavre.
46:01:910Paolo Guiotto: Des moisra vra.
46:04:770Paolo Guiotto: formula.
46:06:290Paolo Guiotto: For the roots, huh?
46:08:220Paolo Guiotto: So you have something like Z to the N equals W,
46:13:440Paolo Guiotto: You want to find out Zedda?
46:15:770Paolo Guiotto: these are the end routes, and apart for W,
46:20:600Paolo Guiotto: equals 0, where there is a unique root, so let's take W different from 0. This equation has N exactly n solutions.
46:29:650Paolo Guiotto: And these are obtained in this way, Z is equal to… to see the solutions, we need to write W in the trigonomatic form, so W equals raw E to I
46:41:840Paolo Guiotto: theta, let's say. Now, the solution is this. Z is a number that, in trigonometric form, has modulus, the root of the modulus of W, so rho to 1 over n.
46:56:380Paolo Guiotto: And then we have an exponential of I. We take the argument, we divide by the order of the root, n, and we add the multiples of 2 pi divided by n. This is the formula for the solution, the Moabra formula.
47:14:400Paolo Guiotto: Now, we have to do this, the point is that we need to represent the right-hand side, which is this one, in this case, in trigonometric form, and this is different for K positive and negative, because these points here
47:29:250Paolo Guiotto: the points,
47:31:990Paolo Guiotto: iK2 pi, no? For k equals 0, this is the value, we already discussed it. For k positive, they are, like here, I2 pi.
47:42:580Paolo Guiotto: I for pi, I6 pi, and so on. For K negative, they are here, minus i2 pi.
47:51:240Paolo Guiotto: minus I for phi, and so on.
47:54:280Paolo Guiotto: And what changes the argument? Because for these ones, the argument is pi half.
48:00:320Paolo Guiotto: But for these ones, the argument is 3 half pi.
48:04:590Paolo Guiotto: So that's why we need to distinguish K positive from K negative.
48:08:850Paolo Guiotto: So, for K positive, I have that I, K2 pi.
48:16:250Paolo Guiotto: or I2 pi k. The modulus of this number is…
48:25:790Paolo Guiotto: do not complicate things. Why? I have my number is I something, so you read the… that says the ordinate of the point, yeah, she says zero.
48:36:150Paolo Guiotto: So all this, at the end, will produce 2 pi K.
48:39:540Paolo Guiotto: It's just this modulus. It's the distance of this to the other. The argument we said is pi half, so e to i pi half.
48:49:30Paolo Guiotto: Yeah.
48:55:730Paolo Guiotto: Okay, so this is the trigonometric form of that number. So Z square equal to this means that Z is equal to the root of 2 pi k
49:10:910Paolo Guiotto: And then we have RE to i. We have to take that angle, the more of a formula, divide by the order of the root, which is 2 here. So take, if you want, pi half, which is the theta, and divide by 2, then add the multiples of 2 pi divided by the order of the root, which is 2.
49:30:600Paolo Guiotto: And all this goes into the argument of this number.
49:35:250Paolo Guiotto: So these are root of 2 phi k.
49:40:980Paolo Guiotto: E to i pi over 4 plus multiples of pi.
49:47:540Paolo Guiotto: And, we know that, the diff… different… Sorry, sorry, sorry, I did the math because I used the K for the 2, but let's use a different letter here. Let's use J.
50:00:810Paolo Guiotto: Sorry.
50:03:430Paolo Guiotto: Because K here is involved with this number, so let's use a different index, J. And how many roots are there? For square roots, there are two roots, so the values are J equals 0 or 1.
50:18:310Paolo Guiotto: So you start always from zero.
50:20:820Paolo Guiotto: So these Z, solutions of E to…
50:26:470Paolo Guiotto: Z squared equals 1, for the K is K positive are these numbers ZK, that for each K there are two numbers.
50:36:820Paolo Guiotto: So, if you want, you can see in this way, root over 2 pi k, then when j is 0, you get E2I pi over 4.
50:48:500Paolo Guiotto: This is the unitary number on the… on the circle, on the unitary single, with the angle pi over 4, which is 45 degrees.
50:57:710Paolo Guiotto: And the two coordinates for this, the sine of… so we should write sine of pi over 4 plus i cosine pi over 4.
51:09:110Paolo Guiotto: These two numbers are root of 2 over 2 plus i root of 2 over 3, yeah.
51:16:120Paolo Guiotto: I'm really… so this is cosine of this plus i sine. But nothing changed for this case, because the values are always equal to sine.
51:27:750Paolo Guiotto: To root of 2 over 2. So we have this solution.
51:32:550Paolo Guiotto: for j equals 0. For j equals 1, the angle is pi over 4 plus pi, so this is the angle, and so the two coordinates are similar, 2 pi k, but now we have just the opposite, minus root of 2 over 2.
51:49:410Paolo Guiotto: plus I minus root of 2 over 2.
51:55:50Paolo Guiotto: This is the ZK for K positive.
51:58:410Paolo Guiotto: Oh, yeah.
52:04:20Paolo Guiotto: Yeah, this is the, the MOIVR formula. It says that
52:08:690Paolo Guiotto: the nth roots of number W are exactly n numbers, that's the… so there are two square roots.
52:19:790Paolo Guiotto: 3 cubic suits, 4 fourth roots, and so on. And they are given by this formula. If W is written in trigonometric form, raw equals raw time C i theta, so, or if you prefer.
52:33:760Paolo Guiotto: raw cosida… plus I sine theta.
52:41:570Paolo Guiotto: then the trigonometric form of the roots is the following. The modulus is the nth root of the modulus of W, so you take W to 1 over N.
52:52:160Paolo Guiotto: then you have the angle that you have to put as the angle into the exponential is made in this way. You take the principal argument that you have there, the theta, you divide by the order of the root, then you take multiples, and that's the J, multiples.
53:09:510Paolo Guiotto: of 2 pi over n. So the total angle, 360 degrees, divided into n parts.
53:16:530Paolo Guiotto: So, what happens? That this, in principle, should be any integer, so in Z.
53:23:310Paolo Guiotto: But you see that when you start from 0, you get 0, 1, 2. When you arrive at N,
53:30:660Paolo Guiotto: Here, it happens that N over N cancels, so this angle is B split 2 pi.
53:37:970Paolo Guiotto: So geometrically, it is the same angle, and this means that this number belongs to the unitary circle with downward. This plus 360 degrees, so it's the same angle.
53:50:290Paolo Guiotto: So you get the same as for J equals 0.
53:53:980Paolo Guiotto: And when a j is n plus 1, you get the same as for J equals 1, and so on. So there are only…
54:00:780Paolo Guiotto: to have distinct solutions, N values, so they are from 0 to the order of the root minus 1, because we have different values. So, in particular, if you have the square root n is 2, so this goes from 0 to 2 minus 1, 1. So these are the two values.
54:21:30Paolo Guiotto: Okay? Now, this is for the KK po… for the case K positive. Then there is a K… K is K negative.
54:31:810Paolo Guiotto: Where I25K…
54:36:440Paolo Guiotto: Now, you see that I cannot say that the modulus is 2 pi k, because this would be negative now.
54:43:180Paolo Guiotto: The number is down here, so in this case, I have I2 pi k.
54:49:900Paolo Guiotto: K is negative, so this distance is actually the absolute value of 2 pi k, or if you want, minus 2 pi k is the same. You don't have to be scared, because there is the minus. The minus, yes, but the k is negative, so that's a positive number.
55:06:600Paolo Guiotto: And the angle here is 3 half pi, so 3 half pi.
55:12:750Paolo Guiotto: So, when we do the roots of this, because Z square is this.
55:17:950Paolo Guiotto: So Z square equal I2 by K,
55:22:440Paolo Guiotto: means Z equal the roots of the modulus, which is minus 2 pi k. Since k is negative, I repeat, remind that K is negative here.
55:34:50Paolo Guiotto: Exponential of the principal argument divided by the order of the root, which is 2, because we are doing the square root, so this is 3 fourth pi.
55:44:330Paolo Guiotto: plus multiples of the total angle 2 pi divided by the order of the root, which is 2. So again, multiples of pi, which J for J equals 0, 1, because there are two roots.
55:59:870Paolo Guiotto: And you can see that these are more or less the same numbers as before, because the angles are 3 fourths pi, which is this one.
56:11:120Paolo Guiotto: And then we have for the… this for J equals 0. For J equals 1, we have to do 3 fourths pi plus another pi. Where do we go?
56:20:710Paolo Guiotto: So if you are here, and we do plus 180 degrees, we go back here.
56:26:780Paolo Guiotto: So the two angles are for J equals 0, You get a 3-4 pi.
56:32:850Paolo Guiotto: And for J equals 1, you can say pi over 4.
56:37:190Paolo Guiotto: Because it's the same angle for this trigonometic patches.
56:40:930Paolo Guiotto: And therefore, here you can write the solutions as above, okay.
56:46:670Paolo Guiotto: Okay, let's take a break. 10 minutes.
57:13:260Paolo Guiotto: Yes.
57:29:550Paolo Guiotto: Yeah, that's good.
58:00:910Paolo Guiotto: Included.
58:48:690Paolo Guiotto: I don't know.
58:51:280Paolo Guiotto: No.
58:52:610Paolo Guiotto: Probably what we need to do.
58:54:900Paolo Guiotto: Beautiful.
58:57:800Paolo Guiotto: What's the name?
58:58:800Paolo Guiotto: Not according to the Lord.
59:04:940Paolo Guiotto: It wouldn't work.
59:11:40Paolo Guiotto: What's up?
59:20:990Paolo Guiotto: Maybe, like, three, you're gonna go up with him.
59:26:260Paolo Guiotto: Beautiful.
59:35:480Paolo Guiotto: Very well.
59:51:760Paolo Guiotto: Okay.
00:15:100Paolo Guiotto: That's ocean is still exposing.
00:22:240Paolo Guiotto: Alright.
00:46:480Paolo Guiotto: That's the song, regular.
00:48:570Paolo Guiotto: Yeah.
01:08:80Paolo Guiotto: Right there, anyway.
01:23:840Paolo Guiotto: What's over 13.
01:29:630Paolo Guiotto: But that's the start of… You don't want to try to say anything.
01:47:150Paolo Guiotto: That'd be the Okay, it's okay.
02:08:530Paolo Guiotto: What do you know?
02:13:980Paolo Guiotto: Yeah, invisible.
02:23:520Paolo Guiotto: Where else?
02:33:390Paolo Guiotto: Thank you.
02:41:240Paolo Guiotto: That's true.
02:48:140Paolo Guiotto: I mean, that's true.
02:59:350Paolo Guiotto: To discuss delivery.
03:01:350Paolo Guiotto: And… Yeah, yeah, yeah, so we made it for that point.
03:07:660Paolo Guiotto: I'm not so…
03:45:600Paolo Guiotto: So in the first year…
04:03:20Paolo Guiotto: Okay, thanks, Roxins.
04:18:560Paolo Guiotto: Okay, so now, let's introduce…
04:21:980Paolo Guiotto: the important definition for which we are doing these functions of complex variables, the definition of C differentiability.
04:40:920Paolo Guiotto: or C derivative.
04:43:150Paolo Guiotto: So now the point is that we can introduce the definition of derivative for a function of complex variable exactly in the same way we do for functions of real value.
04:54:50Paolo Guiotto: variable real value. So we do not need to do all this mess we have done for,
05:01:520Paolo Guiotto: the differentiability of functions of vector, variable, vector, valid. This is because of a simple reason. Here we can…
05:10:530Paolo Guiotto: we have a perfectly working algebraic structure. So, let F be a function of complex variable.
05:20:890Paolo Guiotto: defined on Sun Domain D, that would be now a subset of the complex plane.
05:26:840Paolo Guiotto: We said, bet.
05:32:520Paolo Guiotto: F… is C. Differentiable.
05:40:80Paolo Guiotto: At some point, zedda?
05:46:780Paolo Guiotto: Domain D.
05:49:420Paolo Guiotto: If… There exists the limit
05:56:240Paolo Guiotto: when, sorry, H goes to zero.
06:02:960Paolo Guiotto: of the standard definition of derivative. I take F of Z plus H minus F of Z, so
06:11:160Paolo Guiotto: An increment of the function.
06:13:540Paolo Guiotto: between Z and Z plus H, with H, which is going to 0, and I divide by the incremental divider, which is just the number H. Notice that here, I don't need to justify anything, because
06:27:120Paolo Guiotto: Numerator is the difference between complex numbers, so it's a complex number, denominator is a complex number, and division between complex numbers is always defined, except when the denominator is zero, which is not the case here, because H is going to zero, and as usual, it means that it is different from zero.
06:47:510Paolo Guiotto: So, if this limit exists, and it is, of course, a complex number, and the value of the limit is what we call the derivative of F at point Z.
06:57:840Paolo Guiotto: We call just the derivative.
07:00:190Paolo Guiotto: Now, we see this definition on a few examples.
07:05:830Paolo Guiotto: So, some examples.
07:11:840Paolo Guiotto: exactly as we do when we study the derivative for functions of real variables. So, for example, if the function is constantly equal to a value C, that now is a complex number.
07:25:150Paolo Guiotto: What is the derivative? There exists always the derivative of F, and it will be equal to 0, as you may expect.
07:33:690Paolo Guiotto: So, derivative of constants is 0 also for this definition. We just check this, because…
07:44:40Paolo Guiotto: We just write the definition, the limit when H goes to 0 of F of Z plus H
07:52:230Paolo Guiotto: minus F of Z divided by H,
07:56:950Paolo Guiotto: Well, this numerator is 0, because F is constant, so this guy is C, this guy is C, C minus C is 0, so we are doing the limit when H goes to 0 of 0 divided by H. But 0 divided by any number is 0. So we are doing the limit of 0, we will get 0.
08:17:550Paolo Guiotto: Second example, we computed the derivative of a power.
08:24:29Paolo Guiotto: for the moment, with the integer exponent z to the n.
08:29:229Paolo Guiotto: with N is a natural number. So, to make it significant, of course, n will be greater or equal than 1.
08:37:410Paolo Guiotto: Now, what do you expect? Exactly what happens with the power for real variables. So, there exists a derivative of F, and it is equal to n times z to n minus 1, piece for every Z in C.
08:54:609Paolo Guiotto: So, as you can see, the derivative seems to work in the same way, at least on the algebraic functions we know.
09:02:529Paolo Guiotto: So also here we can, see the calculation that justifies this.
09:08:600Paolo Guiotto: Because if we do the limit.
09:11:210Paolo Guiotto: When H goes to 0 of F of Z plus H.
09:16:470Paolo Guiotto: minus F of Z.
09:19:100Paolo Guiotto: divided by H.
09:23:770Paolo Guiotto: Now,
09:26:330Paolo Guiotto: If we replace the definition of the function here, limit when H goes to 0 of the function is the nth power, so it will be Z plus H to power n minus Z to the n divided always by H.
09:43:300Paolo Guiotto: Now, to compute Z plus H to power n, we use the binomial expansion, or the Newton formula. It's the same, the different names for the same thing. So this is the formula to say A plus B to the N. It's an algebraic stuff that says this is the sum for k going from 0 to N.
10:03:860Paolo Guiotto: You have the binomial coefficient n over k, A to K, B to N minus k. This is the formula.
10:11:230Paolo Guiotto: Okay?
10:12:970Paolo Guiotto: Or…
10:15:710Paolo Guiotto: maybe we… we… we just… it's the same formula, but I want to switch the indexes here, so N over K, but A to N minus K, B to K.
10:28:90Paolo Guiotto: Okay.
10:29:230Paolo Guiotto: So I use this second one that says… so this is the sum for k going from 0 to n, the denomial coefficient n over k, then we have a Z to N minus k times H to K.
10:43:520Paolo Guiotto: Now, if we look at the first few terms of this.
10:47:230Paolo Guiotto: So let me continue here. This is N over 0 Z to the K equals 0, Z to the NH to 0 is 1.
10:55:890Paolo Guiotto: plus N over 1Z to N minus 1H.
11:01:780Paolo Guiotto: And then we have something like n over 2, Z to N minus 2, H squared, and so on, okay?
11:11:460Paolo Guiotto: But look at this, the first coefficients, n over k is n factorial divided k factorial times n minus k factorial, right?
11:23:220Paolo Guiotto: So n over 0 is n factorial over 0 factorial times n minus 0, so n factorial. 0 factorial, as you know, is 1 by definition.
11:36:310Paolo Guiotto: And the n over n makes 1. So this coefficient is 1.
11:41:800Paolo Guiotto: And the other we need is n over 1, which is n factorial divided 1 factorial times n minus 1 factorial.
11:52:300Paolo Guiotto: 1 factorial is 1,
11:54:800Paolo Guiotto: And n factorial divided by n minus 1 factorial, we can rewrite this as n times n minus 1 factorial. So we cancel this, and we get N. So this one is equal to N.
12:08:930Paolo Guiotto: So when we do this expansion, we get this formula. So, Z plus H to power n is equal to… so the first term is n over… n over 0, the binomial coefficient, which is 1, times Z ton, so Z2N.
12:28:540Paolo Guiotto: The next term is the binomial coefficient n over 1, which is n times z to n minus 1 times H, so right, n z to n minus 1 times H, plus
12:41:530Paolo Guiotto: The next coefficient and the next terms, I do not need to expand particularly. This is n over 2, Z to N minus 2 times H squared plus etc, let's say.
12:53:780Paolo Guiotto: So as you can see, when I move this on the other side, I get the numerator of this limit. You see? This one is Z plus H to power n minus Z to the n, which is exactly what I obtained by carrying that Z to the n to the other side. So I can say that
13:13:230Paolo Guiotto: So, the limiter, when H goes to zero of Z plus H to the N,
13:21:70Paolo Guiotto: minus Z to the n divided by H, is equal to the limit when H goes to 0. When I carry this on the other side, what remains is n z to n minus 1 times H, and then the other terms, like n over 2,
13:40:130Paolo Guiotto: Z to N minus 2H squared, and etc. All this must be divided by H.
13:47:850Paolo Guiotto: So when I do the division, I get the limit when H goes to 0. The first term is this divided by H. As you can see, this cancels H, and we get N to… z to n minus 1.
14:01:900Paolo Guiotto: Plus, the next is n over 2, Z ton minus 2, then we have h squared divided the H, so it remains an H.
14:11:610Paolo Guiotto: In the next one, that is N over 3, z to n minus 3, then I have h cubed divided H, so it will remain H squared, and so on.
14:23:190Paolo Guiotto: The last one is N over N. It would be Z to N minus n, so Z to 0, 1. H2N divided by H, it is H to N minus 1. This is the form.
14:38:870Paolo Guiotto: And now, if you look at this, this part is independent of age, so it's a constant.
14:44:400Paolo Guiotto: All these parts contains H that is going to zero, and all these factors are going to zero. So, all this block here is a big block that goes to zero.
14:58:490Paolo Guiotto: So what remains is NZ2N minus 1. This for every Z in C, and so this is the calculation
15:08:110Paolo Guiotto: of what? Of this limit, which is the derivative of F. So this shows that.
15:14:830Paolo Guiotto: there exists the derivative of F, of Z,
15:20:530Paolo Guiotto: And it is equal to n times z to n minus 1, as claimed.
15:26:20Paolo Guiotto: Okay.
15:27:940Paolo Guiotto: Now, if we want to expand our, let's say, our,
15:34:890Paolo Guiotto: our library of derivatives, we should compute, for example, derivatives of polynomials.
15:42:390Paolo Guiotto: A polynomial is a linear combination of powers, so since we know how to derive powers, maybe we can use some general rule that we can get exactly in the same way that you have, so I will not repeat the proof. So, these are the algebraic
16:01:130Paolo Guiotto: rules, off.
16:04:510Paolo Guiotto: calculus.
16:06:320Paolo Guiotto: So it says that, huh?
16:15:710Paolo Guiotto: What, sorry? Yeah.
16:23:290Paolo Guiotto: Yeah, F of Z was the function Z to power n.
16:27:100Paolo Guiotto: So we proved that this function has the derivative, and the derivative is n times z to exponent n-1. So this says that if, if…
16:38:130Paolo Guiotto: F and G are two functions, the C differentiable.
16:43:710Paolo Guiotto: at some point, Zed, Bam.
16:47:880Paolo Guiotto: Number one, also, there's some… their difference is differentiable.
16:54:740Paolo Guiotto: is C differentiable at point Z. And of course, you will have the classical formula. What is the derivative of the sum? What is the derivative of the difference? It is the sum or the difference of the derivatives.
17:10:300Paolo Guiotto: You see?
17:12:710Paolo Guiotto: Then… What about the algebraic product?
17:16:680Paolo Guiotto: The algebraic product is C-differentiable.
17:22:500Paolo Guiotto: At point 7. And what is the derivative?
17:26:690Paolo Guiotto: The derivative is not the product of the derivatives, as you know, that's… the formula we have is the same formula we have for the real derivative. So we do the derivative of one factor, and we multiply… I'm writing Z, but it is… I'm writing X, but it is Z.
17:43:380Paolo Guiotto: So, F prime z times G of Z.
17:47:220Paolo Guiotto: plus F of Z times G prime of Z.
17:51:150Paolo Guiotto: So the same formula you know.
17:53:670Paolo Guiotto: We can do also. The derivative of the ratio is… C differentiable.
18:00:550Paolo Guiotto: Of course, provided the ratio makes sense.
18:04:50Paolo Guiotto: That's Z, then.
18:08:620Paolo Guiotto: Provided… The denominator is different from 0, so G of Z is different from 0.
18:16:660Paolo Guiotto: And that's exactly as it happens for the real derivative, and the formula is the same. What is the derivative of the ratio?
18:24:430Paolo Guiotto: is not the ratio of the derivatives, but it's a much more complicated formula. It's a fraction where the denominator is the square of the denominator, and the numerator is similar to the derivative of the product, but with minus, so we have
18:38:790Paolo Guiotto: Derivative of numerator times denominator minus numerator times the derivative of denominator.
18:48:290Paolo Guiotto: And that… these are the formulas.
18:51:730Paolo Guiotto: From these formulas, it follows some particular consequences.
18:58:860Paolo Guiotto: In particular, which is useful to keep in mind, we have that when we do a linear combination of two functions, the derivative of a linear combination… so linear combination means that
19:14:210Paolo Guiotto: Alpha and beta are not functions, are coefficients, are numbers. So, alpha and beta are numbers, that here, since we are dealing with complex numbers, they can be complex numbers.
19:26:250Paolo Guiotto: Now, the linear… the derivative of the linear combination is the linear combination of the derivative. So, this will be alpha F prime z plus beta G prime Z.
19:40:280Paolo Guiotto: So if you want even more particular, if you do the derivative of alpha F, this is alpha f prime, okay? So you can carry outside.
19:49:600Paolo Guiotto: the constants from the derivative. And the second particular formula is that if G is different from 0, you can have a special case of the derivative of the ratio, which is actually the derivative of the reciprocal of a function G,
20:06:670Paolo Guiotto: And this is minus G prime Z over GZ squared.
20:12:580Paolo Guiotto: That's the formula for the derivative of duration.
20:17:120Paolo Guiotto: Okay, so from this, we have, we can enlarge, so, from this formula, from…
20:29:460Paolo Guiotto: from these… rules.
20:34:900Paolo Guiotto: We have… For example, if we now consider a polynomial.
20:41:470Paolo Guiotto: So, if P of Z is a function like C0 plus C1Z plus C2Z squared, and so on, let's say that it is an nth degree polynomial.
20:55:850Paolo Guiotto: CNZ to DN. We can look at this as a combination of things, so a constant plus constant times a first degree polynomial, sorry, first degree power, constant times second degree power, constant times first degree power, and so on, constant times
21:15:450Paolo Guiotto: And degree power, so we can apply the linearity, which is this property here, linearity.
21:25:700Paolo Guiotto: and say that the derivative of the polynomial will be the sum of the derivatives. Derivatives of C0, C0 is a constant, is 0 plus
21:37:560Paolo Guiotto: Derivative of C1Z, if you want, derivative of C2Z squared, plus derivative of C3Z cubed, etc, CNZ to DN.
21:50:560Paolo Guiotto: But now, the derivative is linear, so you can carry outside the constants.
21:55:720Paolo Guiotto: Here, so we get C1 times the derivative of Z. This is a power, it's the power Z to exponent 1.
22:04:880Paolo Guiotto: No? It's this one, and we say that the derivative is 1Z to 1 minus 1Z to 0, which is, for us, equal to 1.
22:13:600Paolo Guiotto: So we can say that this is C1 times 1 plus C2 times 2Z plus C3 times 3Z squared, and so on, plus CN times NZ2N minus 1.
22:29:450Paolo Guiotto: So, we get in particular that polynomials are differentiable functions.
22:34:160Paolo Guiotto: And, for example, an important class of functions is the class of rational functions, so a ratio between two polynomials, PZ over Qz.
22:46:530Paolo Guiotto: where P and Q are polynomials.
22:54:480Paolo Guiotto: So, we can say that, if,
22:58:70Paolo Guiotto: Q of Z is different from zero. We apply the ratio formula, and we have that P of Z divided Q of Z can be differentiated, and it is… denominator will be the square of Q of Z, which is, of course, a polynomial.
23:16:320Paolo Guiotto: being Q a polynomial. Then we have P prime of Z, which is a polynomial, times Q of Z, which is another polynomial, minus P of Z times Q prime of Z. So it's again… so this is a rational
23:34:480Paolo Guiotto: function.
23:36:970Paolo Guiotto: And this is, again, a rational.
23:40:570Paolo Guiotto: function.
23:44:950Paolo Guiotto: Now, to extend… to really extend the class of differentiable function, the next step is to consider the so-called polynomial of infinite degree, so what technically I'll call the power series.
23:59:40Paolo Guiotto: Now, here we have a nice result. The proof is technical, so I will omit entirely here.
24:04:690Paolo Guiotto: So this is the proposition.
24:08:510Paolo Guiotto: If, F of Z
24:12:20Paolo Guiotto: is the sum of a power series, sum for k going from 0 to infinity, CKZ to K.
24:21:100Paolo Guiotto: is a power.
24:25:680Paolo Guiotto: series, and this result is particularly important because we know that the most important elementary function, exponential, hyperbolic function, trigonometric function.
24:38:330Paolo Guiotto: logarithm is different, but these, at least these five functions are of this type, okay, and many others.
24:46:860Paolo Guiotto: So, it's a positive… it's a power series with
24:51:980Paolo Guiotto: Positive radius, otherwise the theorem is false, with the radius of convergence, are positive.
25:04:80Paolo Guiotto: Then, what turns out is, we know that this radius of convergence basically tells that
25:11:340Paolo Guiotto: Inside the disk of convergence, you have convergence. Outside, you have not convergence. So, inside the disk centered at 0 with radius R, this series is well-defined and convergent.
25:24:660Paolo Guiotto: But it turns out that it is also differentiable, so there exists the derivative of F.
25:31:420Paolo Guiotto: a poinsettia.
25:33:200Paolo Guiotto: for every Z in the disc of convergence, so for every Z with the models less than R, and the derivative can be computed by formally doing the derivative term by term.
25:45:670Paolo Guiotto: So, the point is that imagine that you want to differentiate that sum as if it is a polynomial. What have we done here for the polynomial? We said we want to differentiate that, it's a derivative of a sum, we do the sum of the derivatives.
26:01:200Paolo Guiotto: Right? This is possible because we have a rule here that tells the derivative of the sum is the sum of the derivatives. This rule holds for 2, but as you can imagine, you can iterate, and it holds when you have to sum 3, 4, 1 million of functions, but still always a finite number of functions.
26:20:570Paolo Guiotto: So when you pass to this case, these are infinitely many terms, so you… it's not easy to say that the derivative of the sum is the sum of the derivatives. That's non-trivial. So you cannot just say, we use linearity, and derivative of the sum is sum of the derivative.
26:37:50Paolo Guiotto: This is true when the sum is finite, but when the sum is infinite, this becomes a little bit more complicated. But for this case, it's true. So it means that we could differentiate this term by term, huh?
26:50:70Paolo Guiotto: Now, since the term with k equals 0 is a constant, when you differentiate, this disappears, and it remains terms from k from 1 to infinity, and the formula is that if you differentiate them, you get K, CK, Z to K
27:08:520Paolo Guiotto: minus 1.
27:10:00Paolo Guiotto: Now, K here ranges from 1
27:13:630Paolo Guiotto: to infinity, because k equals 0 is washed out by the derivative, it's a constant.
27:20:700Paolo Guiotto: So, in particular, for example, important example…
27:26:650Paolo Guiotto: If you take the exponential function, it's a power series, sum for k going from 0 to infinity, z to k over k factorial.
27:36:730Paolo Guiotto: And this has radius of convergence, as we know, equal to plus infinity. So, for every Z in C, this is well-defined. Now, by this proposition, it turns out that there exists the derivative of this function.
27:50:960Paolo Guiotto: And this, for every Zed, modulus of Z less than plus infinity, so this means for every Z,
27:58:590Paolo Guiotto: So the exponential is always differentiable.
28:03:630Paolo Guiotto: And what happens when we do the derivative by using the formula you see above in green?
28:09:390Paolo Guiotto: So, we cut off the first term because it's a constant, so the derivative kills this. So we start from 1 to infinity. Now, the CK is 1 over k factorial, that's the CK.
28:22:950Paolo Guiotto: Then we have the term that comes deriving this thing, so this K times Z to K minus 1, right?
28:30:750Paolo Guiotto: But now, we can slightly rearrange this, because this becomes sum for k going from 1 to infinity. So, writing the k factorial down here as, sorry, k times…
28:45:900Paolo Guiotto: K minus 1 factorial, in such a way that I can simplify that K.
28:53:10Paolo Guiotto: You see that this K is up here, and we got the sum for k going from 1 to infinity of 1 over k minus 1 factorial.
29:02:300Paolo Guiotto: times Z to K minus 1.
29:05:170Paolo Guiotto: Now, we can relabel this index here as, for example, J.
29:11:610Paolo Guiotto: No? What happens? Since k is going from 1 to infinity, when you subtract 1, J, which is k minus 1, goes from 0 to infinity. So J goes from 0 to infinity. 1 over
29:27:420Paolo Guiotto: K minus 1 is what we call the J, 1 over J factorial, and here we have Z to J.
29:34:180Paolo Guiotto: And what is this?
29:38:460Paolo Guiotto: it's the exponential series again. So, we got this, that the derivative of the exponential is the exponential. Something that you already know, of course, for the real case, but it told also for the complex case, so for every Z in C.
29:54:320Paolo Guiotto: So we have this remarkable fact. The derivative of the exponential is still the exponential for every Z.
30:02:230Paolo Guiotto: in SEEP.
30:05:30Paolo Guiotto: And similarly, you can do with all the other functions. I will illustrate only the case of the hyperbolic function.
30:13:510Paolo Guiotto: then you can work out all the others. So, for example, remind that hyperbolic sign was also a,
30:25:270Paolo Guiotto: power series, okay? So it is a sum. In this case, we do the sum of odd powers, something like 2K plus 1 over Z… of a 2K plus 1 factorial.
30:40:210Paolo Guiotto: when k goes from 0 to infinity, right? This is the… and the radius of convergence here is also equal to plus infinity, so this thing is always convergent. So, according to the theorem we just state here.
30:56:410Paolo Guiotto: For power series, they are always differentiable sense.
31:02:60Paolo Guiotto: in the interior disk of convergence, and here, since R is equal plus infinity means the interior complex plane, so there exists the derivative of the hyperbolic sine of Z,
31:15:260Paolo Guiotto: and it is equal to… we can do the derivative term by term. Notice that here, since the first power, that one for k equals 0, is Z to 1, there is not the degree 0 term.
31:28:800Paolo Guiotto: So, we keep K equals 0,
31:32:450Paolo Guiotto: to infinity. When we do the derivative, well, this is a coefficient. I write here, 1 over 2K plus 1 factorial. Then I have to differentiate the power. This is…
31:45:560Paolo Guiotto: the exponent, 2K plus 1 times Z to the exponent, decrease it by 1, so it is to 2K.
31:54:70Paolo Guiotto: Again, as you can see here, we can cancel this with part of this, because if we write this as 2K plus 1,
32:02:610Paolo Guiotto: This is the factorial, so it is the product of all integers between 1 and that number.
32:09:40Paolo Guiotto: So you have 1 times 2 times 3 times 4, etc, until you arrive to that one, but just before you have 2K. And then you have 2K minus 1, 2K minus 2, etc, until you go down to 3, 2, 1. But all this is the factorial of 2K.
32:28:320Paolo Guiotto: So you can cancel this with that, and what you get is sum.
32:34:680Paolo Guiotto: for k going from 0 to infinity of Z exponent 2K divided by 2K factorial. What is this?
32:48:220Paolo Guiotto: It is the hyperbolic cosines.
32:52:650Paolo Guiotto: Okay.
32:53:900Paolo Guiotto: And also, this happens with the real case, no? The reality of the hyperbolic sign is the hyperbolic cosine, and similarly.
33:08:690Paolo Guiotto: The derivative of the hyperbolic cosine Is the hyperbolic sign.
33:16:910Paolo Guiotto: You'll proceed in the same way.
33:19:470Paolo Guiotto: And also, example, please check details, do the details. You can differentiate the trigonometric sign, and you get the trigonometric cosine as function of complex variable.
33:33:590Paolo Guiotto: and do the derivative of cosine, and you get minus sine this for every Z in C. So the relations you already know for real still are valid for complex numbers.
33:48:890Paolo Guiotto: Okay, so far so good. We have seen examples of functions which are differentiable, okay?
33:57:450Paolo Guiotto: Now, let's see an example of a function which is not differentiable.
34:02:260Paolo Guiotto: And here, the story starts to become different from the real story. Now, when…
34:09:110Paolo Guiotto: Let's see. When you have to imagine to a function which is not differentiable for the real functions, you imagine functions where there is not the tangent, so functions like this, no?
34:21:200Paolo Guiotto: Imagine a function with an angle there. There is not the tangent there. The typical example that usually is done in first-year calculus is the absolute value.
34:31:300Paolo Guiotto: The absolute value is not differentiable at zero. So, the idea is that it is not differentiable because there is something, there is a kink in the graph, there is something that happens, the function is not regular, is not smooth.
34:45:680Paolo Guiotto: But here it can happen something really different. So let's see this example. This is apparently a very simple example. Take this function, f of z, defined as the real part of Z.
35:02:860Paolo Guiotto: Okay?
35:04:200Paolo Guiotto: So, it's a well-defined function. I have a number Z, I associate its real path. It's real, yes, but I can't see any real number into this big set of complex numbers. So, this is a function that is defined from C to C, okay? So I think always also real as complex numbers.
35:23:920Paolo Guiotto: Well, what turns out, if you want to see analytically how this function works, you could say, if the number Z is given in the form X plus AY, where X and Y are real, the real and imaginary part of your number, this function gives the value X.
35:42:870Paolo Guiotto: So this is how it's defined.
35:44:600Paolo Guiotto: So you see, it's… we may say it's basically the first less trivial function we may think after the constant, a function which is constant, or a function that takes Z and gives you Z, okay?
35:59:120Paolo Guiotto: F of Z equals Z is another simple function. This one does something similar, takes Z and gives you the real part of Z.
36:06:540Paolo Guiotto: But the point is that F is never… C, differentiable.
36:17:570Paolo Guiotto: So this function is never… there is no point where it is differential. We can check, showing that the limit that defines the derivative does not exist.
36:27:750Paolo Guiotto: Indeed, if we compute the limit when H goes to 0 of F of Z plus H,
36:37:350Paolo Guiotto: minus F of Z divided by H. Let's see what happens.
36:43:700Paolo Guiotto: This is the limit when H goes to 0 of the function, takes a number, and gives the real part. So, real part of Z plus H minus real part of Z divided by H.
36:58:990Paolo Guiotto: Now, what is the real part of Z plus H?
37:04:480Paolo Guiotto: Well, the real part is innocent. If you sum two numbers, the real part is the sum of the real parts. So, real part of Z plus real part of H.
37:14:740Paolo Guiotto: So this allows to simplify with the other real part there. So we have to compute the limit for H going to zero of real part of H divided by age. Now, the crucial point is that this limit does not exist.
37:37:320Paolo Guiotto: And let's see why.
37:38:920Paolo Guiotto: And it's like a problem of computing a limit at zero for a function of two variables. We illustrate that there are ways to go to zero along which we get something, and there are ways going to zero along which we get something else.
37:54:210Paolo Guiotto: For example, imagine that here our H is moving in the complex plane. So imagine that I move H just on the real axis, and I send it to zero.
38:05:260Paolo Guiotto: So this means that my age is the number X plus I0, right?
38:10:740Paolo Guiotto: So, this limit, limit when H goes to zero, of real part of H, divided by H, becomes when do you go to zero with this point? What should happen?
38:25:310Paolo Guiotto: This is the point x0, when it goes to zero.
38:29:460Paolo Guiotto: When?
38:31:70Paolo Guiotto: Yes, X goes to 0, that's right. When X goes to zero. So we transform this into the limiter. When x goes to zero of 1. A real part of H is X.
38:42:500Paolo Guiotto: H is X itself, because X plus I0 is just X. So the limit is easy, it comes 1. So it seems simple. But what if we do the limit on this other side? So we move now H on the imaginary part.
38:59:220Paolo Guiotto: So this means that this H is now something like 0 plus IY.
39:05:160Paolo Guiotto: So, the limit when H goes to zero of real part of H divided by H becomes the limit when now it will be…
39:17:260Paolo Guiotto: Who is going to zero?
39:21:180Paolo Guiotto: Well, Y, that's right, Y goes to zero. What is the real part of H?
39:28:460Paolo Guiotto: It is zero. So you have 0 divided, what is HIY? So what is this limit? 0.
39:36:540Paolo Guiotto: And what happens? This limit is 0, this limit is 1. So it means that there are ways to go to 0 along which that ratio goes to 1, other ways along which it goes to 0. So this is just the symptom of… there is no limit for
39:54:730Paolo Guiotto: H going to 0 of real part of H divided by H,
40:00:380Paolo Guiotto: But this means that there is no limit of the incremental ratio, that f of z plus h minus Fz divided H, and this, at the end means there is no derivative.
40:12:580Paolo Guiotto: There is no F prime of Z, and this is independent of Z, as you have seen, for every Z in C. So this function is never differentiable.
40:23:250Paolo Guiotto: And this is quite disturbing or interesting, because, depending on how you look at these things, because you may expect that such a stupid function should be differentiable. It's just one of the coordinates, but it is not.
40:37:840Paolo Guiotto: Now, do the same check exercise, Seth.
40:45:350Paolo Guiotto: That's… the following functions.
40:56:90Paolo Guiotto: R… Never.
40:59:900Paolo Guiotto: differentiable. You have to do the same kind of check we have seen here. So one is the imaginary part of Z, as you may expect, then another one is, for example, the square of the modulus of Z, so that one that is X squared plus Y squared.
41:16:690Paolo Guiotto: How is possible that this is not differentiable? This is not differentiable. Another one, interesting, is the conjugator.
41:25:160Paolo Guiotto: So that one that takes Z and gives you Z conjugate, so X minus IY. So check that these three functions are never differentiable.
41:35:140Paolo Guiotto: Okay.
41:38:950Paolo Guiotto: Have a nice weekend.
41:41:760Paolo Guiotto: And…