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Would you mind explaining how you got those answers please?Analytic in the unit disc
$z^7$ yes
$|z|$ no
$\frac{1}{z}$ yes
Correct?
RV based on that. I shouldn't get the same g=f but so what is wrong with this:$z^{7}$ is complex differntiable everywhere, $|z|$ is complex differentiable nowhere, and $\frac{1}{z}$ is not complex differentiable at the origin
so the last one is incorrect
I know that I think about it. I can't do that since |z| is the modulus not the real the absolute value correct?I was interpreting $|z|$ as the complex conjugate for some reason. Sorry.
I would use the Cauchy Riemann equations to check if $|z|$ is differentiable inside the unit disc.
Do you know how I can solve that integral then?Yeah. $|z| = |x+iy| = \sqrt{x^{2}+y^{2}} = \sqrt{x^{2}+y^{2}} + i 0 = u(x,y) + iv(x,y) $
And the function is differentiable only where the Cauchy-Riemann equations are satisfied:
$u_{x} = \frac{x}{\sqrt{x^{2}+y^{2}}} $ $u_{y}= \frac{y}{\sqrt{x^{2}+y^{2}}}$, $v_{x}=0$, and $v_{y}=0$
$u_{x} = \frac{x}{\sqrt{x^{2}+y^{2}}} = v_{y} = 0$ and $u_{y} = \frac{y}{\sqrt{x^{2}+y^{2}}} = - v_{x} = 0$
The equations are not satisifed at the origin, which is inside the unit disc.
So $|z|$ is not analytic inside the unit disc.
EDIT: Actually, $|z|$ doesn't appear to be differentiable anywhere.