Thread: Using polar form of Cauchy-Riemann equations

1. Using polar form of Cauchy-Riemann equations

How would I go about using the Cauchy-Riemann equations to show that $f(z) = z^n$ is an entire function for any n using polar coordinates?

2. Re: Using polar form of Cauchy-Riemann equations

Well for starters, what is f(z) in polar form? What is the polar form of the Cauchy-Riemann equations?

3. Re: Using polar form of Cauchy-Riemann equations

$f(z) = r^ne^{i\theta n}$

Cauchy-Riemann is $u_r = \frac{1}{r}v_{\theta}$ and $v_r = \frac{-1}{r}u_{\theta}$

I split f(z) up like so:

$u(r,\theta) = r^n \cos(\theta n)$

$v(r,\theta) = r^n \sin(\theta n)$

I'm stuck on what to do next... am I allowed to differentiate as I would over R?

4. Re: Using polar form of Cauchy-Riemann equations

Originally Posted by jgv115
$f(z) = r^ne^{i\theta n}$

Cauchy-Riemann is $u_r = \frac{1}{r}v_{\theta}$ and $v_r = \frac{-1}{r}u_{\theta}$

I split f(z) up like so:

$u(r,\theta) = r^n \cos(\theta n)$

$v(r,\theta) = r^n \sin(\theta n)$

I'm stuck on what to do next... am I allowed to differentiate as I would over R?
$u_r(r,\theta)=n r^{n-1}\cos(n\theta)$
$u_\theta(r,\theta) = -n r^n \sin(n\theta)$