Evaluate complex integral using deformation theorem It's given that, $\gamma$ is a circle of radius $r$ and centre $z=a$ inside $\Gamma$. I need to use the deformation theorem to evaluate this: $$\oint

Evaluate complex integral using deformation theorem It's given that, $\gamma$ is a circle of radius $r$ and centre $z=a$ inside $\Gamma$. I need to use the deformation theorem to evaluate this: $$\oint_\Gamma \ (z-a)^ndz , n\in\mathbb{Z}$$ Surely at different $n$ the answer will be different? I think the integral about $\gamma$ is: $$\oint_{0}^{2\pi} \ (z-a)^n dz$$ but the answer will be different for each integer n? I think that when n is negative the function is not analytic in the countour hence I don't think I can use Cauchy's integral formula or Cauchy's theorem. Also when n is 0 or positive i think the integral is 0 because of its analyticity on the contour Sorry if my assumptions don't make sense, i only started the course, Thanks for any help

Think of $z=a+r e^{i \theta}$. Then $dz=i r e^{i \theta}$ and the contour integral becomes

$$i r^{n+1} \int_0^{2 \pi} d\theta \, e^{i (n+1) \theta}$$

So long as $n$ is an integer, the above integral vanishes...except when $n=-1$, when it is equal to $i 2 \pi$. Yes, even when $n$ is less than $-1$, the integral vanishes.

This is why the residue of a function is defined as the coefficient of the $(z-z_0)^{-1}$ term in the Laurent expansion about a pole $z=z_0$. All other terms in the series vanish upon integration about a closed loop around the pole.