Use residues to find $\oint_C \dfrac{dz}{z^4 + 4}$, where $C$ is the circle $|z + 1 - i| = 1$.
There is a discrepancy in the solution to this problem between my work and the solution.
My Work
$z^4 + 4 = (z^2 - 2i)(z^2 + 2i) = (z - (1 + i))(z - (-1 - i))(z - (i - 1))(z - (1 - i))$
We have simple poles at $\pm(1 + i)$ and $\pm(i - 1)$. However, only the pole $i - 1$ lies inside $C$.
As I understand it, the residue formula is $Res(f, z_0) = \dfrac{1}{(k - 1)!} \lim_{z \to z_0} \dfrac{d^{k - 1}}{dz^{k - 1}} [(z - z_0)^k f(z) ]$
So we have $Res(f, i - 1) = \dfrac{1}{0!} \lim_{z \to i - 1} \left[(z - (i - 1))^1 \dfrac{1}{(z - (1 + i))(z - (-1 - i))(z - (i - 1))(z - (1 - i))}\right]$
$= \lim_{z \to i - 1} \dfrac{1}{(z - (1 + i))(z - (-1 - i))(z - (1 - i))}$
$= \dfrac{1}{(-2)(2i)(2i - 2)} = \dfrac{1}{8 + 8i}$
Therefore, the residue is $\dfrac{1}{8 + 8i}$.
And so, using Cauchy's residue theorem, we have that $\oint_C \dfrac{dz}{z^4 + 4} = 2\pi i \left( \dfrac{1}{8 + 8i} \right)$
However, the solution has the residue of $\dfrac{1}{z^4 + 4}$ at $z_0 = -1 + i$ as $\dfrac{1 - i}{16}$. This is obviously different from my solution. Even so, I cannot find any errors in my reasoning. After all, we both have the same pole of $-1 + i$, and, as far as I can tell from checking my textbook, my formula for the residue is correct?
I would greatly appreciate it if people could please take the time to review my work and explain why this discrepancy exists. Is my work incorrect? Or is it the case that the solution is incorrect?
$$ \dfrac{1}{8 + 8i} = \dfrac{1-i}{8(1 + i)(1-i)} = \dfrac{1-i}{8(1 -i^2)} = \dfrac{1-i}{16} $$
so it's all fine!