Complex derivative confusion: z^2 = (x+iy)^2 = (|z|^2 e^{2i*theta})

[Mondo]

Hello,

I consider the derivative of [imath]z^2 = (x+iy)^2 = (|z|^2e^{2i\theta})[/imath] when I calculate it with respect to [imath]z[/imath] I get [imath]2z[/imath] this means that the length of the vector [imath]|z|[/imath] is duplicated and there is a rotation by an angle [imath]\theta[/imath]. In other words, if I have a number [imath]z[/imath] and want to get it's image after an infinitesimal move I need to multiply it's length by [imath]2|z|[/imath] and rotate by angle [imath]\theta[/imath]. However when I calculate this same derivative with respect to [imath]\theta[/imath] I get [imath]d/d\theta [e^{2i\theta}] = 2ie^{2i\theta}[/imath] quite different result. Why?

Thank you.

 

[topsquark]

Hello,

I consider the derivative of [imath]z^2 = (x+iy)^2 = (|z|^2e^{2i\theta})[/imath] when I calculate it with respect to [imath]z[/imath] I get [imath]2z[/imath] this means that the length of the vector [imath]|z|[/imath] is duplicated and there is a rotation by an angle [imath]\theta[/imath]. In other words, if I have a number [imath]z[/imath] and want to get it's image after an infinitesimal move I need to multiply it's length by [imath]2|z|[/imath] and rotate by angle [imath]\theta[/imath]. However when I calculate this same derivative with respect to [imath]\theta[/imath] I get [imath]d/d\theta [e^{2i\theta}] = 2ie^{2i\theta}[/imath] quite different result. Why?

Thank you.

You need to clarify something first. Are you talking about taking a complex derivative? Or are you simply taking partials wrt x and y? (See * below.)

We can show, using the Cauchy-Riemann equations that [imath]z^2[/imath] does indeed have a derivative. (You always need to show this before you start taking derivatives.)

Further, you need to explain what you mean by [imath]z^2[/imath]. Taken according to definition this is [imath]z^*z[/imath] and when you take the derivative you indeed get
[imath]\dfrac{d}{dz} z^* z = \left ( \dfrac{ \partial }{ \partial x } + i \dfrac{ \partial }{ \partial y } \right ) [ (x - iy)(x + iy) ] = 2(x + iy) = 2z[/imath].

But note that, the literal expression [imath]z^2 = z \cdot z[/imath] has
[imath]\dfrac{d}{dz} z^2 = \left ( \dfrac{ \partial }{ \partial x } + i \dfrac{ \partial }{ \partial y } \right ) [ (x + iy)(x + iy) ] = 0[/imath].

When you took your z derivative of [imath]z^2 = (x+iy)^2 = (|z|^2e^{2i\theta})[/imath] you weren't nearly careful enough. At this stage you should convert everything to x and y and take the partials that way (once you've determined that you can take the derivative in the first place.)

Now, as to the meaning of the derivative d/dz on a function F(z). It has the same kind of meaning that we ascribe to d/dx on a function y(x): it is the instantaneous rate of change of z on F(z) at a given point. However, you can't draw neat things like slopes unless you can sketch in 4D. I think the best way to look at it is via the first order approximation:
[imath]F(z + a) = F(a) + \left . \dfrac{dF}{dz} \right |_{z = a} \cdot (z - a)[/imath]

Note that everything here is a complex number, so saying that the derivative "moves" the function in a certain direction no longer can be done.

(*) Now, unless I mistook your meaning and you are simply looking at partials of the magnitude of z and the angle, then d/dz tells you how the magnitude is changing and [imath]d/d\theta[/imath] tells you how the angle is changing, ie. how the "vector" z is rotating in the Argand plane. Talking about how 2z is how the "length of the vector is duplicated and there is a rotation by an angle" is confusing. If you are looking at z as a vector in the Argand plane, then you need to keep your components distinct, just like you keep x and y distinct.

-Dan

 

[blamocur]

However when I calculate this same derivative with respect to θ\thetaθ I get d/dθ[e2iθ]=2ie2iθd/d\theta [e^{2i\theta}] = 2ie^{2i\theta}d/dθ[e2iθ]=2ie2iθ quite different result.

Why do expect the derivatives with respect to [imath]z[/imath] and with respect to [imath]\theta[/imath] to be equal?

 

[Mondo]

@topsquark, thank you for your answer. However in many respects I don't agree. For instance, why do you define [imath]z^2 = z*z = (x-iy)(x+iy)[/imath] this is not a square of [imath]z[/imath] but rather a product of [imath]z[/imath] and its conjugate - two completely different things.

When you took your z derivative of z2=(x+iy)2=(∣z∣2e2iθ)z^2 = (x+iy)^2 = (|z|^2e^{2i\theta})z2=(x+iy)2=(∣z∣2e2iθ) you weren't nearly careful enough. At this stage you should convert everything to x and y and take the partials that way (once you've determined that you can take the derivative in the first place.)

Why should I convert to cartesian format? Polar format is just as good, if we wrote in a polar form we have a length part and a rotation part which is exponential function of the angle. So my hope was to get the same answers when differentiating WRT to z and angle but for some reason they are not the same.

@blamocur, as I said above - there is only one complex number that we square, namely [imath]z = (x+iy) = e^{i\theta}[/imath]. For ease of calculation let's assume it a complex number on the unit circle so it's length is unity. Now I want to calculate it's derivative I do it first WRT to z and get [math]\frac{d}{dz}z^2 = 2z[/math] which means the derivative is [imath]2e^{i\theta}[/imath] and then WRT to angle [imath]\frac{d}{d\theta}(e^{2i\theta}) = 2ie^{2i\theta}[/imath] a completely different result while I would expect to get the same answer right? I you say no, why not?

 

[topsquark]

@topsquark, thank you for your answer. However in many respects I don't agree. For instance, why do you define [imath]z^2 = z*z = (x-iy)(x+iy)[/imath] this is not a square of [imath]z[/imath] but rather a product of [imath]z[/imath] and its conjugate - two completely different things.

Why should I convert to cartesian format? Polar format is just as good, if we wrote in a polar form we have a length part and a rotation part which is exponential function of the angle. So my hope was to get the same answers when differentiating WRT to z and angle but for some reason they are not the same.

@blamocur, as I said above - there is only one complex number that we square, namely [imath]z = (x+iy) = e^{i\theta}[/imath]. For ease of calculation let's assume it a complex number on the unit circle so it's length is unity. Now I want to calculate it's derivative I do it first WRT to z and get [math]\frac{d}{dz}z^2 = 2z[/math] which means the derivative is [imath]2e^{i\theta}[/imath] and then WRT to angle [imath]\frac{d}{d\theta}(e^{2i\theta}) = 2ie^{2i\theta}[/imath] a completely different result while I would expect to get the same answer right? I you say no, why not?

The symbol [imath]z^2[/imath] is often used to express [imath]z^*z[/imath]. But [imath]z^2 = z \cdot z[/imath] is not how you were treating the expression when you tried to take the derivative.

Why did I convert to Cartesian form? So I could get the correct answer!

As I pointed out above
[imath]\dfrac{d}{dz} z^2 = 0[/imath].

Tell me where I go wrong:
[imath]\dfrac{d}{dz} z^2 = \left ( \dfrac{ \partial }{ \partial x } + i \dfrac{ \partial }{ \partial y } \right ) (x + iy)^2[/imath]

[imath]= \left ( \dfrac{ \partial }{ \partial x } + i \dfrac{ \partial }{ \partial y } \right ) (x^2 - y^2 + 2ixy)[/imath]

[imath]= \dfrac{ \partial }{ \partial x } (x^2 - y^2 + 2ixy) + i \dfrac{ \partial }{ \partial y } (x^2 - y^2 + 2ixy)[/imath]

[imath]= (2x + 2iy) + i (- 2y + 2ix)[/imath]

[imath]= 2x + 2iy - 2iy - 2x = 0[/imath]

Definitely not 2z as the power rule would imply.

Complex derivatives do not follow the same rules as real derivatives.

This site is littered with ads, but it does the job.

-Dan

 

[Mondo]

The symbol z2z^2z2 is often used to express z∗zz^*zz∗z. But z2=z⋅zz^2 = z \cdot zz2=z⋅z is not how you were treating the expression when you tried to take the derivative.

This is the first thing I don't agree with - why do you claim z time conjugate of z is a square of z?

Why did I convert to Cartesian form? So I could get the correct answer!

This doesn't make sense to me - ok, so you claim if I multiply two complex numbers in a rectangular form I will get right answer but if I do the same in say polar form I won't? This can not be true. One representation may be better or worse suited for a particular job but both must produce the right answer. So maybe the way I calculate the derivative in polar form is incorrect.. that's what I try to figure out.

Tell me where I go wrong:

You multiply by 'i' for I don't know what reason.

Also, tell me where do I go wrong?
[imath]z^2 = (x+iy)^2 = x^2 - y^2 + 2ixy -> u=x^2-y^2; v = 2ixy[/imath]
Now calculation of partials reveals
[imath]\partial_x(u) = 2x; \partial_y(u) = -2y; \partial_x(v) = i2y; \partial_y(v) = i2x[/imath]
hence [imath]\frac{d}{dz}z^2 = 2x - 2y + i(2x + 2y) \ne 0[/imath]

 

[blamocur]

...while I would expect to get the same answer right? I you say no, why not?

I've already asked you why you expect to get the same answer, and I haven't seen any explanation. "Why not?" is not a way to justify, let alone to prove, your statements.

 

[topsquark]

This is the first thing I don't agree with - why do you claim z time conjugate of z is a square of z?

This doesn't make sense to me - ok, so you claim if I multiply two complex numbers in a rectangular form I will get right answer but if I do the same in say polar form I won't? This can not be true. One representation may be better or worse suited for a particular job but both must produce the right answer. So maybe the way I calculate the derivative in polar form is incorrect.. that's what I try to figure out.


You multiply by 'i' for I don't know what reason.

Also, tell me where do I go wrong?
[imath]z^2 = (x+iy)^2 = x^2 - y^2 + 2ixy -> u=x^2-y^2; v = 2ixy[/imath]
Now calculation of partials reveals
[imath]\partial_x(u) = 2x; \partial_y(u) = -2y; \partial_x(v) = i2y; \partial_y(v) = i2x[/imath]
hence [imath]\frac{d}{dz}z^2 = 2x - 2y + i(2x + 2y) \ne 0[/imath]

[imath]z^*z[/imath] is not the "time conjugate." (Where did you get the "time" from?) It's just the conjugate. And, again, I commonly see it represented as [imath]z^2[/imath]. Bear in mind I read a lot of Physics. But we can drop that: if you are calling [imath]z^2 = z \cdot z[/imath] then I don't have a problem with that, just so long as we all know what is being referenced.

Now for my "Oh, crap!" moment. I wasn't completely wrong above: I didn't specify which path I was using to take the derivative and wound up with the wrong answer. But I wasn't careful enough. The trouble with complex derivatives is that you have to make sure that the path you are using to take the derivative doesn't affect the derivative. (That's how the Cauchy-Riemann conditions are derived.) So, if I modified what I did as letting y = f(x):
[imath]\dfrac{d}{dz} = \dfrac{ \partial }{ \partial x} + i \dfrac{ \partial }{ \partial y} = \dfrac{ \partial }{ \partial x} + i \dfrac{ \partial }{ \partial (f(x))}[/imath]
(which involves an extra df/dx from the chain rule) it would have come out right. No one in their right mind would do it this way if they could avoid it: they just choose a path along the x or y axis.

My apologies for the confusion!

Okay, so I have to re-address your OP.

I'm going to echo blamocur's comment: Why should the z and [imath]\theta[/imath] derivatives be the same?

-Dan

 

[Mondo]

I've already asked you why you expect to get the same answer, and I haven't seen any explanation. "Why not?" is not a way to justify, let alone to prove, your statements.

@blamocur now? I gave extensive answer to your question - I calculated a derivative of the same complex number in rectangular and polar form and showed that it gives two different results while it should give the same since again, it is a derivative of the same thing! In the same way as when you multiply two complex numbers in different representations you get the same product. How can I better justify my expectations?

 

[blamocur]

@blamocur now? I gave extensive answer to your question - I calculated a derivative of the same complex number in rectangular and polar form and showed that it gives two different results while it should give the same since again, it is a derivative of the same thing! In the same way as when you multiply two complex numbers in different representations you get the same product. How can I better justify my expectations?

You've calculated a derivative of the same complex function, but over different variables, i.e., [imath]z[/imath] and [imath]\theta[/imath]. I.e. there is more than one derivative possible when there is more than one argument/variable. The fact that [imath]z[/imath] is complex and [imath]\theta[/imath] is real only adds to the confusion.

To simplify things you can look at multi-variate functions in real domain, for example [imath]f(x,y) = x^2 + 2y[/imath]. If you look at [imath]\frac{\partial f}{\partial x}[/imath] and [imath]\frac{\partial f}{\partial y}[/imath] you don't expect them to be the same everywhere (e.g. x=1,y=1).

 

[blamocur]

You've calculated a derivative of the same complex function, but over different variables, i.e., [imath]z[/imath] and [imath]\theta[/imath]. I.e. there is more than one derivative possible when there is more than one argument/variable. The fact that [imath]z[/imath] is complex and [imath]\theta[/imath] is real only adds to the confusion.

To simplify things you can look at multi-variate functions in real domain, for example [imath]f(x,y) = x^2 + 2y[/imath]. If you look at [imath]\frac{\partial f}{\partial x}[/imath] and [imath]\frac{\partial f}{\partial y}[/imath] you don't expect them to be the same everywhere (e.g. x=1,y=1).

Oops: the derivatives are the same at (1,1), but they are different everywhere except the [imath]x=1[/imath] line.

 

[Mondo]

@blamocur but here is the thing: in polar form we have two distinct information encoded, a vector length and angle it makes with the positive real axis. For instance [imath]|z|e^{i\theta}[/imath] tells you right away two things, it is a complex number with a modulus, length of [imath]|z|[/imath] and an angle [imath]\theta[/imath]. So it seems to be logical to me, that when I have a function of a complex number in a polar form say [imath]f(z) = |z|^2e^{2i\theta}[/imath] to calculate derivatives of it separately, namely one for the change of the length and the other for the angle. Why can I do it here? Because they don't depend on each other. IOW, no matter how much I change the vector length it's angle it makes with positive real axis will always be the same and vice versa, no matter how much I rotate a vector its length won't be affected right? So where is my mistake?

 

[blamocur]

@blamocur but here is the thing: in polar form we have two distinct information encoded, a vector length and angle it makes with the positive real axis. For instance [imath]|z|e^{i\theta}[/imath] tells you right away two things, it is a complex number with a modulus, length of [imath]|z|[/imath] and an angle [imath]\theta[/imath]. So it seems to be logical to me, that when I have a function of a complex number in a polar form say [imath]f(z) = |z|^2e^{2i\theta}[/imath] to calculate derivatives of it separately, namely one for the change of the length and the other for the angle. Why can I do it here? Because they don't depend on each other. IOW, no matter how much I change the vector length it's angle it makes with positive real axis will always be the same and vice versa, no matter how much I rotate a vector its length won't be affected right? So where is my mistake?

I don't see any mistakes in this post. You are correct to say that any function [imath]f(z)[/imath] of complex variable looked at as a function of two real variables [imath]f(|z|, \theta)[/imath] (although I'd prefer to write [imath]f(r,\theta)[/imath]). Just as you can express it as [imath]f(x,y)[/imath], where [imath]z = x+iy[/imath].

What I don't see in any of your posts is an attempt to show how the partial derivative over [imath]\theta[/imath] is related to the ordinary derivative over [imath]z[/imath].

Here is a hint: look at the chain rule where [imath]f[/imath] is a function of [imath]z[/imath], and [imath]z[/imath] is a function of [imath]r[/imath] and [imath]\theta[/imath] :
[math]\frac{\partial f}{\partial \theta} = \frac{df}{dz} \frac{\partial z}{\partial \theta}[/math]

 

[Mondo]

@blamocur, the mistake I made is in this sentence of my previous post

Why can I do it here? Because they don't depend on each other

In fact this angle depends on x and y - I fooled myself in thinking they don't because in the polar form it looks it is the case but I think the catch is that in the polar form you can only do two operations, namely extend the vector and rotate it. And both of them keep the same ratio between x and y.