Exercise on page 518

Mondo

No. that is incorrect.

Vasco

So here is the problem I have with these complex derivatives and it already popped out a few times in this thread: $f(z) = z^2 = (x+iy)^2 = (|z|e^{i\theta})^2$ so we have here three ways to represent the same complex number/function. Now, when differentiating these formulas I would expect to get the same derivative, regardless of the representation of a number in the same way as it doesn't make any difference if I multiply two numbers in cartesian or polar form.

Actually it is not that easy to show that multiplication in rectangular and polar form gives the same answer: for our $z^2$ example, we have $(x^2-y^2 + 2ixy)$ in rectangular form and $|z|^2e^{i\theta}$ in a polar form. If we would like to compare vector lengths after the mapping then, the polar form gives the answer right away $|z|^2$ but the rectangular way gives ugly looking result $\sqrt{(x^2-y^2)^2 + (2xy)^2}$ but it actually can be simplified to $x^2+y^2$ (exercise )
But going back to the original problem $z = x+iy$ so it seems to be natural to define the derivative of $z$ as $(d/dx(x) + id/dy(y))$
I will read chapter 4 again as I haven't done fully yet. Not sure I will find an answer for that there though.

Mondo

No. $\nabla=\partial_x+i\partial_y$ but $\nabla$ is not $\frac{d}{dz}$

Vasco

Mondo

You need to understand the derivation of (6) and (7) on page 210. I have said this several times before. If you don't fully understand how these results are obtained you will not be able to go forward.

Vasco

I meant of $z$ with respect to $z$ on the LHS and WRT $(x,y)$ on the RHS.



I reread that and I think I understand it. The problem seems to be that in the way I proposed I don't check the change in $v$ caused by a change in $x$ which is wrong. CR equations calculate partials of both $u,v$ in regards to both $x,y$ and a relation between them is summarized in equations (7) and (8). So for $z^2 = (x^2 - y^2 + 2ixy = u + iv)$ the derivative WRT $x$ is $2x + i2y = 2e^{i\theta}$ (assuming unit circle). Next, WRT to $y$ -> $-2y +i2x = -i\frac{d}{y}z^2$, So it shows that the derivative of $f(z)$ is $\partial_x(u) + i\partial_x(v)$.
However what about my previous consideration of $f(z) = |z|^2e^{2i\theta}$ here the exponential part doesn't have any dependency with the vector length - no matter what is the vector length it will always be rotated the same amount, by the same angle $\theta$. So it seems logical to consider derivatives of vector length and rotation separately in polar form. Why it doesn't work Vasco?

Yes I found this document very helpful users.math.msu.edu/users/shapiro/teaching/classes/425/crpolar.pdf which basically shows that my assumption:

is incorrect and angle $\theta$ actually does depend on $(x,y)$. This can be seen at least from $e^{i\theta} = cos(x) + isin(x)$. So I fooled myself in thinking vector length and it's angle are "independent". Do you agree with this Vasco?
Also do you agree with my conclusion from the previous post?