Calculating a complex integral with generalized Cauchy integral formula

Question

Calculate the integral $$\oint\limits_{|z|=2} \frac{z^m}{(1-z)^n}\mathrm{d}z, \, \text{for}\ n,m\in \mathbb{N}$$

So in fact of the denominator I think I have to use the generalized Cauchy integral formula. I got this: $$\oint\limits_{|z|=2} \frac{z^m}{(1-z)^n}\mathrm{d}z = \frac{(n-1)!}{2\pi \mathrm{i}}f^{(1)}(n-1)=\frac{(n-1)!}{2\pi \mathrm{i}}\cdot m (n-1)^{m-1}$$

Is this the answer? That seems to easy, but the pole is inside the circle and $z^m$ is a holomorphic differentiable function. So I think there is no problem, isn't it? Thank you!

Answer 1

Of course you will get the result if you apply the Cauchy formula correctly, $$ \oint_{|z|=2}\frac{f(z)}{(z-1)^n}dz=\frac{2\pi i}{(n-1)!}f^{(n-1)}(1). $$

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Alternatively, you can apply the binomial theorem to $$ z^m=(1+(z-1))^m $$ to find $$ \frac{z^m}{(z-1)^n}=\sum_{k=0}^m\binom{m}k(z-1)^{k-n} $$ where the only term giving a non-trivial integral is with $k-n=-1$.

Answer 2

The $n$'th derivative of $z^m$, $m\ge n$ is given by

$$\frac{d^n z^m}{dz^n}=\frac{m!}{(m-n)!}z^{m-n}$$

Therefore, we have

$$\begin{align} \oint_{|z|=2} \frac{z^m}{(1-z)^n}\,dz&=2\pi i \frac{(-1)^n \,m!}{(n-1)!(m-n+1)!}\\\\ &=2\pi i (-1)^n\,\binom{m}{n-1} \end{align}$$