Show that $f(z) =z^n$ where $n$ is a positive integer, is analytic
Find its derivative.
I tried solving it using Cauchy Riemann equation. But for that, $f(z)$ needs to be separated as $f(z)=u+iv$.
$f(z)=(x+iy)^n$
We can expand this binomially then group the real and imaginary parts and then take their partial derivatives. But that would be a tedious process.
I am not able to split $f(z) =z^n =u+iv$. Please help.
Best Answer
It's much simpler than that. If $z_0\in\mathbb C$, then$$z^n=\bigl(z_0+(z-z_0)\bigr)^n=\sum_{k=0}^n\binom nk{z_0}^{n-k}(z-z_0)^k.$$This expresses $z^n$ as the sum of a power series centered at $z_0$.