Show that $f(z) =z^n$ where $n$ is a positive integer, is analytic .

Question

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.

Answer 1

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 $z_0$.

Answer 2

An analytic function is a function that is locally given by a convergent power series, so it's trivial. I am assuming you mean, $f$ is differentiable everywhere. In this case, as in the real case, it is just matter of evaluating $\lim_{z\to z_0}\frac{z^{n}-z^{n}_0}{z-z_0}.$

Answer 3

The direct definition of the derivative is straight forward.

$$ \lim_{z\to z_0} \frac {z^n-z_0^n}{z-z_0} =\lim _{z\to z_0} (z^{n-1} +z^{n-2}z_0 +...+z_0^{n-1}) = nz_0^{n-1}$$