Find the residue of $f(z)$ at $z=i$ Find the residue of the function $f(z)$ at $z=i$
$$f(z) = \frac{Log\ z}{(z^2+1)^2} $$
So I did this $$= \frac{Log\ z}{(z+i)^2(z-i)^2}$$
Then $$\frac{Log\ z}{(z+i)^2}\Biggr|_{z=i} = \frac{Log\ i}{(2i)^2} = \frac{Log\ i}{-4}$$
I don't know what else to do with this. The book says the answer should be $\frac{\pi + 2i}{8}$
 A: I assume that you mean to take the principal argument for $\log$; that is, you're writing
$$
\operatorname{Log} z := \log\left|z\right| + i\operatorname{Arg}(z),
$$
where $\operatorname{Arg}(z)$ is the principal argument for $z$ (the one that takes values in $(-\pi,\pi]$).
Then
$$
\operatorname{Log} i = \log(1) + i\operatorname{Arg}(i) = 0 + i\pi/2.
$$
Hence, $f(z)$ has a pole of order $2$ at $z = i$. Recall that if $\alpha$ is a pole of $f$ of order $n$, then we may compute the residue of $f$ at $\alpha$ as
$$
\operatorname{Res}(f,\alpha) = \frac{1}{(n-1)!}\lim_{z\to\alpha}\frac{d^{n-1}}{dz^{n-1}}\left[(z - \alpha)^n f(z)\right].
$$
Thus,
\begin{align*}
\operatorname{Res}(f,i) &= \lim_{z\to i}\frac{d}{dz}\left[(z - i)^2 f(z)\right]\\
&=\lim_{z\to i}\frac{d}{dz}\left[\frac{\operatorname{Log}z}{(z + i)^2}\right]\\
&= \lim_{z\to i}\frac{\frac{(z + i)^2}{z} - 2(z +i)\operatorname{Log}z}{(z + i)^4}\\
&=\lim_{z\to i}\frac{\frac{(z + i)}{z} - 2\operatorname{Log}z}{(z + i)^3}\\
&= \frac{\frac{2i}{i} - 2\operatorname{Log}i}{(2i)^3}\\
&= \frac{2 - i\pi}{-8i}\\
&= \frac{(i\pi -2)(-i)}{8}\\
&= \frac{\pi + 2i}{8}.
\end{align*}
