Calculating poles and residues of given function

Question

Let $$ f(z) = \frac{1}{(z+i)^7} - \frac{3}{z-i} = \frac{z-i-3(z+i)^7}{(z+i)^7(z-i)} $$ This has pole of order 1 in $i$ and order 7 in $-i$. I can easily calculate the residue from pole at $i$:

$$ \text{Res}(f,i) = \lim_{z \to i} (z-i)\frac{z-i-3(z+i)^7}{(z+i)^7(z-i)} = \frac{z-i-3(z+i)^7}{(z+i)^7} = \frac{-3(2i)^7}{(2i)^7} = -3 $$

I don't know what to do with the second residue, from the pole in $-i$.

The pole is of order 7 so to calculate it I'd have to define function $$ g(z) = (z+i)^7\frac{z-i-3(z+i)^7}{(z+i)^7(z-i)} = \frac{z-i-3(z+i)^7}{(z-i)} $$ calculate its $(7-1)$th derivative and: $$ \text{Res}(f,-i) = \lim_{z \to -i}\frac{g^{(m-1)}(z)}{(m-1)!} $$ But calculating 6th derivative of such function($\frac{z-i-3(z+i)^7}{(z-i)} = 1 - 3\frac{(z+i)^7}{z-i}$) seems like a daunting task. Am I missing something crucial or do I really have to count 6 derivatives of that?

Answer 1

You overcomplicate your calculations.

Note that the residue is $\mathbb{C}$-linear, hence

$$Res\left(\frac{1}{(z+i)^7} - \frac{3}{z-i},z_0\right) = Res\left(\frac{1}{(z+i)^7},z_0\right) - Res\left(\frac{3}{z-i},z_0\right)$$

Since $\frac{1}{(z+i)^7}$ is holomorph at $z=i$ and $\frac{3}{z-i}$ is holomorph at $z=-i$, you get

$$Res(f,i)= -Res\left(\frac{3}{z-i},i\right)=-\lim_{z\to i}\frac{3(z-i)}{z-i}=-3$$

and, if you are still not equipped with Laurent series

$$Res(f,-i)= Res\left(\frac{1}{(z+i)^7},-i\right)= \frac 1{6!}\lim_{z\to -i}\frac{d^6}{dz^6}\frac{(z+i)^7}{(z+i)^7}= 0$$

Answer 2

The Laurent series of $\frac1{(z+i)^7}$ centered at $-i$ is $\frac1{(z+i)^7}$ itself. Therefore, the residue of $\frac1{(z+i)^7}$ at $-i$ is $0$. And $-i$ is a regular point of $\frac3{z-i}$. So, the residue that you're after is $0$.