How to find this integral using Cauchy integral formula

Question

How to obtain that $$\int\limits_{|z|=r} (\bar{z})^{-m} z^{-n-1}\, dz = \begin{cases} 2\pi ir^{-2m} &\text{if}\,\,n=m, \\ 0 &\text{if}\,\,n \neq m, \end{cases}$$ for $r>0$.

I suppose I have to use Cauchy integral formula:

• if $n=m$ then we have $$\int\limits_{|z|=r} (\bar{z})^{-m} z^{-m-1}\, dz=\int\limits_{|z|=r} (\bar{z})^{-m} z^{-m}\, \frac{dz}{z}=\int\limits_{|z|=r} |z|^{-2m} \frac{dz}{z-0}=2\pi i |0|^{-2m}=\begin{cases} 2\pi i, m=0\\ 0,\quad m \neq 0\end{cases}$$ and I am confused.

I could use $z=re^{i\theta}$ to obtain $$\int\limits_{|z|=r} |z|^{-2m} \frac{dz}{z}=\int\limits_{0}^{2\pi}r^{-2m}\, i \, d\theta=2\pi i r^{-2m},$$ but what is wrong with my first way?

• for $n \neq m$ I would do same, and then result would be $2\pi i \bar{0}^{-m}0^{-n}=0$ using Cauchy integral formula.

Answer 1

The reason that your first "approach" is incorrect is that the function $$f(z) = |z|^{-2m}$$ is not holomorphic (except when $m=0$). Cauchy's integral formula is only valid for holomorphic functions.

Answer 2

In your first way, you should have done like that

$\int\limits_{|z|=r}|z|^{-2m}\frac{dz}{z} = r^{-2m}\int\limits_{|z|=r}\frac{dz}{z} = 2\pi ir^{-2m}$, because |z| is constant on the curve |z|=r.

(as coffeemath said)