Power rule derivative in complex Problem:
Prove that if $f(z)= z^n$, then $f' (z)$ = $n z^{n-1} $ using the definition of the derivative.
 A: Just write out: $$f(z+h) = \sum_{k=0}^n \binom{n}{k}z^{n-k}h^k = f(z) + h(nz^{n-1}) + h^2\cdot\sum\dots.$$
A: Suppose $f(z)=z^n.$ Then for all $a,$ we have
$$\begin{align}f'(a) &= \lim_{z\to a}\frac{f(z)-f(a)}{z-a}\\ &= \lim_{z\to a}\frac{z^n-a^n}{z-a}\\ &= \lim_{z\to a}\frac{(z-a)(z^{n-1}+az^{n-2}+a^2z^{n-3}+\cdots+a^{n-2}z+a^{n-1})}{z-a}\\ &= \lim_{z\to a}(z^{n-1}+az^{n-2}+a^2z^{n-3}+\cdots+a^{n-2}z+a^{n-1})\\ &= \underbrace{a^{n-1}+a^{n-1}+\cdots+a^{n-1}}_{n\text{ times}}\\ &= na^{n-1}.\end{align}$$
A: For positive powers, the proof is straightforward using the expansion of $(z+h)^n$ in Jakub Konieczny's answer.
For negative powers, there is a tad more algebra work involved. Suppose $n>0$. We prove $(z^{-n})'=-nz^{-n-1}$:
\begin{align*}\require{cancel}
(z^{-n})' & = \lim_{h\to0}\dfrac{(z+h)^{-n}-z^{-n}}{h} \\
 & = \lim_{h\to0}\dfrac{\dfrac{1}{z^n + nz^{n-1}h+\binom{n}{2}z^{n-2}h^2+\cdots+h^n}-\dfrac{1}{z^n}}{h} \\
 & = \lim_{h\to0}\dfrac{\cancel{z^n-z^n}-nz^{n-1}h-\binom{n}{2}z^{n-2}h^2-\cdots-h^n}{h(z^{2n} + nz^{2n-1}h+\binom{n}{2}z^{2n-2}h^2+\cdots+z^{n}h^n)} \\
 & = \lim_{h\to0}\dfrac{-nz^{n-1}-\binom{n}{2}z^{n-2}h-\cdots-h^{n-1}}{z^{2n} + nz^{2n-1}h+\binom{n}{2}z^{2n-2}h^2+\cdots+z^{n}h^n} \\
 & = \dfrac{-nz^{n-1}}{z^{2n}} \\
 & = -nz^{-n-1} \;. \;\square
\end{align*}
