Prove $\left(z^n\right)' = nz^{n-1}$

Question

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:

1. 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).
2. 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.
3. 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! :-)

Answer 1

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!!!

Answer 2

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} $.