# 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:

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! :-)

• Use the Product Rule and induction. – André Nicolas Sep 23 '14 at 5:43

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 $f'(x) =f(x)\dfrac{n}{x} =x^n\dfrac{n}{x} =n x^{n-1}$.

• Now that is addiction! Best head off to "LDA"; you know what those letters stand for! One might also ask if introducing $\ln x$ for complex $x$ is "Keeping it simple."! But you're derivation is cute, so +1 Step in any event! With the Wisdom to know the difference 'twixt your answer and mine, I say, "My name's Bob, thanks for letting me share!" . . . Cheers! – Robert Lewis Sep 23 '14 at 5:34
• Google gives its first few results for "LDA" as "Latent Dirichlet allocation", "Linear discriminant analysis", and "Learning Disabilities Association". My meaning, of course, is "logarithmic differentiation addiction", and this comment is an effort to make that more widely known. – marty cohen Sep 23 '14 at 5:40
• I figured it would be evident from the context;-)! – Robert Lewis Sep 23 '14 at 5:41
• Not to me. My internal pattern matcher first identified "LDA" as the assembly language mnemonic for "load accumulator", possibly a relic of my 6502 programming days. – marty cohen Sep 23 '14 at 5:44
• Ha! Now that is truly context-free! – Robert Lewis Sep 23 '14 at 5:46

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!!!