Compute $\int_{|z-i|=1}\frac{\log(z+4)}{(z^2+1)^2}$

Question

Compute $\displaystyle\int_{|z-i|=1}\frac{\log(z+4)}{(z^2+1)^2}$

As I see it there are two possibilities:

1. $g(z)=\frac{\log(z+4)}{(z+i)^2}$ is holomorphic on $D_1(i)$, so we have

$\displaystyle\int_{|z-i|=1}\frac{\log(z+4)}{(z^2+1)^2}=\oint_{|z-i|=1}\frac{g(z)}{(z-i)^2}=2\pi i\frac{d}{dz}\left(g(z)\right)\mid_{z=i}$

2. Also we see that $z=i$ is an isolated singularity in $D_1(i)$ with winding number 1, so we have

$\displaystyle\int_{|z-i|=1}\frac{\log(z+4)}{(z^2+1)^2}=2\pi i\text{Res}_i\left(\frac{\log(z+4)}{(z^2+1)^2}\right)$

and since $z=i$ is a pole of order 2 we can obtain the residue via

$\displaystyle \frac{d}{dz}\left((z-i)^2f(z)\right)\mid_{z\to i}=\frac{2\log(z+4)}{(z-i)(z+i)^2}\mid_{z\to i}+\frac{\log(z+4)}{(z+i)^2}\mid _{z\to i}$

Now I see that I must have done something wrong since the first summand is problematic.

Answer 1

Both forms are indeed equivalent. Simply note that

$$\begin{align} \frac{d}{dz}\left((z-i)^2f(z)\right)&=2(z-i)f(z)+(z-i)^2f'(z)\\\\ &=2(z-i)\frac{\log(z+4)}{(z^2+1)^2}+(z-i)^2\left(\frac{1}{(z+4)(z^2+1)^2}-\frac{4z\log(z+4)}{(z^2+1)^3}\right)\\\\ &=\frac{1}{(z+4)(z+i)^2}+\frac{2\log(z+4)}{(z+i)^3}\left(\frac{z+i}{z-i}-\frac{2z}{z-i}\right)\\\\ &=\frac{1}{(z+4)(z+i)^2}-\frac{2\log(z+4)}{(z+i)^3}\\\\ &=\frac{d}{dz}\left(\frac{\log(z+4)}{(z+i)^2}\right) \end{align}$$

as was to be shown!

Answer 2

The problem comes fron the derivative.

$$ \frac{d}{dz}\left((z-i)^2f(z)\right)=\frac{d}{dz}\Bigl(\frac{\log(z+4)}{(z+i)^2}\Bigr). $$