# Analyticity of z

• Apr 29th 2011, 06:50 AM
Glitch
Analyticity of z
I was reading through my text, and it stated that $z^n$ (for some integer n) was not analytic at n = 0. Why is this the case? Usually I'd substitute z = x + iy and then use the Cauchy-Riemann equations to check, but (x + iy)^n makes it difficult to separate the real and imaginary parts.

Thanks.
• Apr 29th 2011, 06:56 AM
TheEmptySet
Quote:

Originally Posted by Glitch
I was reading through my text, and it stated that $z^n$ (for some integer n) was not analytic at n = 0. Why is this the case? Usually I'd substitute z = x + iy and then use the Cauchy-Riemann equations to check, but (x + iy)^n makes it difficult to separate the real and imaginary parts.

Thanks.

Do you mean when $z=0$ not n?
• Apr 29th 2011, 07:05 AM
Glitch
Sorry, I should probably provide some context.

The question I got this from is the following:

Let C be the circle of radius R centre the origin, taken anticlockwise. Find $\int_C{z^n \dz}$ for integer values of n.
• Apr 29th 2011, 07:17 AM
TheEmptySet
Quote:

Originally Posted by Glitch
Sorry, I should probably provide some context.

The question I got this from is the following:

Let C be the circle of radius R centre the origin, taken anticlockwise. Find $\int_C{z^n \dz}$ for integer values of n.

Well you can parametrize the circle at the origin by

$z=Re^{i\theta},\theta \in [0,2\pi] \implies dz=Rie^{i\theta}d\theta$

When

$n=0 \implies z^n=1$ so you get the integral

$\oint dz=Ri\int_{0}^{2\pi}e^{i\theta}d\theta=0$

In fact you can use this for every integer n but be careful with the case n=-1
• Apr 29th 2011, 07:21 AM
Glitch
I just don't understand why it is not analytic at n = 0 (at least, according to this text).
• Apr 29th 2011, 07:34 AM
TheEmptySet
Quote:

Originally Posted by Glitch
I just don't understand why it is not analytic at n = 0 (at least, according to this text).

It should be when n=0 we get

$z^0=\begin{cases}1, z \ne 0 \\ \text{undefined},z=0 \end{cases}$

Technically $0^0$ is undefined, but if we check the limit along any path of

$\lim_{z \to 0}z^0=1$ So by the Riemann removable singularity theorem the function can be analytically extended to all of the complex plane

Removable singularity - Wikipedia, the free encyclopedia

Note that holomorphic function is a function that is analytic on all of the complex plane.