Prove $\left(z^n\right)' = nz^{n-1}$ I'm trying to solve this complex-variable problem:

Prove, using direct Calculus, that $\left(z^n\right)' = nz^{n-1}$ ($n \in \mathbb{N}$).

I tried the following steps to solve that:


*

*I saw that $z^n = \left( x + iy \right)^n = r^n \text{e}^{i\cdot n\theta} = r^n \left[ \cos{(n\theta)} + i\sin{(n\theta)} \right]$ (using Moivre's formula).

*Then, I used these Cauchy-Riemann conditions:
$$\dfrac{\partial U}{\partial r} = \dfrac{1}{r} \dfrac{\partial V}{\partial \theta} \text{ and } \dfrac{\partial V}{\partial r} = -\dfrac{1}{r} \dfrac{\partial U}{\partial \theta}$$
to see if the given expression can be derivated.

*So, using those conditions, and assuming $U(r,\theta) = r^n \cos{(n\theta)}$ and $V(r,\theta) = r^n \sin{(n\theta)}$:
$$\dfrac{\partial U}{\partial r} = nr^{n-1} \cos{(n\theta)} = \dfrac{1}{r} \dfrac{\partial V}{\partial \theta} \cdots (1)$$
$$\dfrac{\partial V}{\partial r} = nr^{n-1} \sin{(n\theta)} = -\dfrac{1}{r} \dfrac{\partial U}{\partial \theta} \cdots (2)$$


As you can see, $(1)$ and $(2)$ indicates the given expression can be derivated, but now I don't know what to do next. I already now that, if I work with functions in $x$ and $y$,
$$f'(z) = \dfrac{\partial U}{\partial x} + i\dfrac{\partial V}{\partial y} = \dfrac{\partial V}{\partial x} - i\dfrac{\partial U}{\partial y}$$
But I tried to do the same with $r$ and $\theta$, and I can't prove this with that way.
Any help will be appreciated. Thanks in advance! :-)
 A: Don't know what is meant by
"direct calculus",
but I will indulge my addiction to
logarithmic differentiation:
Let
$f(x) = x^n$.
Then $\ln f(x) = n \ln x$.
Differentiating,
$\dfrac{f'(x)}{f(x)}
=\dfrac{n}{x}
$,
so
$f'(x)
=f(x)\dfrac{n}{x}
=x^n\dfrac{n}{x}
=n x^{n-1}
$.
A: The most "direct calculus" I know is precisely as in the real case:
Setting
$f(z) = z^n, \tag{1}$
we have
$f(z + \Delta z) = (z + \Delta z)^n = \sum_0^n \dfrac{n!}{j!(n - j)!}z^{n - j}(\Delta z)^j, \tag{2}$
by the binomial theorem; thus
$f(z + \Delta z) - f(z) = (z + \Delta z)^n - z^n = \sum_1^n \dfrac{n!}{j!(n - j)!}z^{n - j}(\Delta z)^j, \tag{3}$
whence
$\dfrac{f(z + \Delta z) - f(z)}{\Delta z} = \sum_1^n \dfrac{n!}{j!(n - j)!}z^{n - j}(\Delta z)^{j - 1}$
$= nz^{n - 1} + \sum_2^n \dfrac{n!}{j!(n - j)!}z^{n - j}(\Delta z)^{j - 1} =  nz^{n - 1} + \Delta z \sum_2^n \dfrac{n!}{j!(n - j)!}z^{n - j}(\Delta z)^{j - 2}. \tag{5}$
For fixed $z$, it is now easy to see that
$\lim_{\Delta z \to 0}\dfrac{f(z + \Delta z) - f(z)}{\Delta z} = nz^{n - 1}, \tag{6}$
since
$\Delta z \sum_2^n \dfrac{n!}{j!(n - j)!}z^{n - j}(\Delta z)^{j - 2} \to 0 \; \; \text{as} \; \Delta z \to 0; \tag{6}$
the above shows, by about as "direct calculus" that there is, that
$(z^n)' = nz^{n - 1}. \tag{7}$
The preceding works for $n \ge 2$; I leave the exceedingly simple cases $n = 0,1$ to my readers; evaluating
$f'(z) = \lim_{\Delta z \to 0}\dfrac{f(z + \Delta z) - f(z)}{\Delta z} \tag{8}$
is nearly trivial in the event $n \le 1$.
QED.
Hope this helps.  Cheers,
and as always,
Fiat Lux!!!
