How to find the residues of $\frac{1}{(z^4+4)^2}$?

Question

How to find the residues of this function? $$\frac{1}{(z^4+4)^2}$$

So far, I found the poles: $z_1=-1-i$, $z_2 = -1+i$, $z_3=1-i$, $z_4=1+i$. I know they are of the second order. But I have troubles with finding the residues.

Answer 1

Example:

The residue at the pole $z=1+i$.

Since $f(z)$ can be written as:

$$f(z) = \frac{1}{(z+1+i)^2(z+1-i)^2(z-1+i)^2(z-1-i)^2}$$

Then

$$\text{res}_{z=1+i}f(z) = \lim_{z\to 1+i} \left(\frac{(z-1-i)^2}{(z+1+i)^2(z+1-i)^2(z-1+i)^2(z-1-i)^2} \right)'$$

$$= \lim_{z\to 1+i} \left(\frac{1}{(z+1+i)^2(z+1-i)^2(z-1+i)^2} \right)'$$

$$= \lim_{z\to 1+i} \left(\frac{1}{\left[(z+1+i)(z^2+2i)\right]^2} \right)'$$

$$= \lim_{z\to 1+i} \frac{-2(z+1+i)(z^2+2i)[z^2+2i+2z(z+1+i)]}{\left[(z+1+i)(z^2+2i)\right]^4}$$

$$=\frac{-2(2+2i)(4i)[4i+(2+2i)(2+2i)]}{[(2+2i)(4i)]^4}$$

Answer 2

There is a lot nicer and more general solution.

You could write the poles as $z=\sqrt{2}e^{i\left(\frac{\pi}{4}+k\frac{\pi}{2} \right)}$ with $k \in \{ 0,1,2,3\}$.

Then $$\text{res}(f,\sqrt{2}e^{i\left(\frac{\pi}{4}+k\frac{\pi}{2} \right)}) = \lim_{z\to \sqrt{2}e^{i\left(\frac{\pi}{4}+k\frac{\pi}{2} \right)}} \left( \frac{\left(z-\sqrt{2}e^{i\left(\frac{\pi}{4}+k\frac{\pi}{2} \right)}\right)^2}{(z^4-4)^2}\right)'$$

After taking the derivative you'd have to use 'l Hospital 3 times to get: $$ = - \lim_{z\to \sqrt{2}e^{i\left(\frac{\pi}{4}+k\frac{\pi}{2} \right)}} \frac{14(z-\sqrt{2}e^{i\left(\frac{\pi}{4}+k\frac{\pi}{2}\right)})+6z}{3(z^4+4)^2+64z^4(z^4+4)+32z^8} $$

$$=\frac{-3}{256}e^{-7i\left(\frac{\pi}{4}+k\frac{\pi}{2}\right)}$$

Which holds for all $k$. (And that's kinda nice :) )