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[SOLVED] Analytic

dwsmith

Well-known member
Feb 1, 2012
1,673
Analytic in the unit disc

$z^7$ yes
$|z|$ no
$\frac{1}{z}$ yes

Correct?
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
Analytic in the unit disc

$z^7$ yes
$|z|$ no
$\frac{1}{z}$ yes

Correct?
Would you mind explaining how you got those answers please?
 

Random Variable

Well-known member
MHB Math Helper
Jan 31, 2012
253
$z^{7}$ is complex differntiable everywhere, EDIT: $\bar{z}$ is complex differentiable nowhere, and $\frac{1}{z}$ is not complex differentiable at the origin

so the last one is incorrect
 
Last edited:

dwsmith

Well-known member
Feb 1, 2012
1,673
$z^{7}$ 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
RV based on that. I shouldn't get the same g=f but so what is wrong with this:

$g(u) = |u|$
Evaluating the series around the unit disc at $z_0 = 0$, we have
$$
f(z) = \frac{1}{2\pi i}\sum_{n = 0}^{\infty}\left[\int_0^{2\pi}\frac{|u|}{u^{n + 1}}du z^n\right].
$$
If we just look at the integral, we have
$$
\sum_{n = 0}^{\infty}\int_0^{2\pi}\frac{|u|}{u^{n + 1}}du =
\begin{cases}
\displaystyle\sum_{n = 0}^{\infty}\int_0^{2\pi} u^{-n} du, & \text{if} \ u\geq 0\\
\displaystyle -\sum_{n = 0}^{\infty}\int_0^{2\pi} u^{-n}du, & \text{if} \ u < 0
\end{cases} =
\begin{cases}
0, & \text{if} \ n\neq 1\\
2\pi i, & \text{if} \ n = 1
\end{cases}
$$
Then $f(z) = z$ if $u\geq 0$ and $f(z) = -z$ if $u < 0$.
Therefore, $f(z) = |z|$.
 

Random Variable

Well-known member
MHB Math Helper
Jan 31, 2012
253
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.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
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.
I know that I think about it. I can't do that since |z| is the modulus not the real the absolute value correct?
 

Random Variable

Well-known member
MHB Math Helper
Jan 31, 2012
253
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.
 
Last edited:

dwsmith

Well-known member
Feb 1, 2012
1,673
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.
Do you know how I can solve that integral then?