Find the residues of singularities of the following function:

Question

Let $$f(z)=\frac{e^{\pi iz}}{z^2-2z+2}$$ and $\gamma_R$ is the closed contour made up by the semi-circular contour $\sigma_1$ given by, $\sigma_1(t)=Re^{it}$, and the straight line $\gamma_2$ from $-R$ to $R$ (a semi circle).

So the singularities of $f$ are $1+i$ and $1-i$ and it's type is 'double'.

The residue theorem is $$\oint f(z)dz=2\pi i\sum_{k=1}^n I(\gamma,a_k)Res(f,a_k).$$

I'm not sure how to use this theorem to calculate the residues though.

Answer 1

The residue of the function inside the contour is

$$\lim_{z \rightarrow 1+i} (z-(1+i)) f(z) $$

$$ = \frac{e^{i \pi (1+i)}}{2 i} $$

The value of the integral is simply $i 2 \pi $ times this residue.

Answer 2

Each pole is a simple pole (not double). The easiest way to compute the residue is

$$\operatorname{Res}(f,1+i) = \frac{e^{\pi i z}}{(z^2-2z+2)'}\bigg|_{z=1+i} = \frac{e^{\pi i(1+i)}}{2(1+i)-2} $$

with a similar formula for the other pole (but the other one is not needed for your integral).

Answer 3

When the numerator is an entire function, the residue at each zero of the denominator is the same as if you just had a constant over the denominator, and the constant is the value of the numerator at that point.

(....or even if the numerator is merely differentiable in some open neighborhood of that point.)