Find principal part of Laurent expansion of $f(z) = \frac{1}{(z^2+1)^2}$ about $z=i$.
My attempt at a solution:
First, I noticed that if I plug in $z=i$, I get a zero in the denominator. This leads me to think that it is an isolated singularity. If I look at the classification of singularities, I believe it is a pole since $\lim_{z \to z_0} \vert f(z_0) \vert = \infty$ for $z_0 = i$.
By recalling the definition of the principal part of $f$, I am looking for the series containing all negative powers of $(z-z_0)$ in the Laurent expansion $\sum_{k=-\infty}^{\infty} a_k(z-z_0)^k$.
Based on what I have seen, I need to find the partial fraction decomposition of $f$. If so, then I have $\frac{1}{(z^2+1)^2} = \frac{A}{z^2+1}+\frac{B}{(z^2+1)^2}$. However, I think I am doing something wrong. From here, I believe I am supposed to use a geometric series.
I am using the textbook Complex Analysis, Third Edition by Joseph Bak and Donald J. Newman.
Any assistance and clarification would be greatly appreciated.
Best Answer
Hint:
Rewrite $f(z)$ as $$ f(z) = \dfrac{1}{(z^2+1)^2}=\dfrac{1}{(z-i)^2(z+i)^2}=\dfrac{-1}{(z-i)^2}\cdot\dfrac{d}{dz}\left(\dfrac{1}{z+i}\right). $$ Using a geometric series, $$ \dfrac{1}{z+i}=\dfrac{1}{z-i+2i}=\dfrac{1}{2i}\cdot\dfrac{1}{\dfrac{z-i}{2i}+1}=\\ =\dfrac{1}{2i}\sum\limits_{n=0}^{\infty}{(-1)^n\left(\dfrac{z-i}{2i}\right)^n} =\dfrac{1}{2i}\sum\limits_{n=0}^{\infty}{\dfrac{(-1)^n}{(2i)^n}({z-i})^n}. $$ Therefore, $$f(z)=\dfrac{-1}{(z-i)^2}\dfrac{d}{dz}\left(\dfrac{1}{z+i}\right)=\dfrac{-1}{(z-i)^2}\dfrac{d}{dz}\left(\dfrac{1}{2i}\sum\limits_{n=0}^{\infty}{\dfrac{(-1)^n}{(2i)^n}({z-i})^n}\right)=\\ =\dfrac{-1}{(z-i)^2}\cdot\dfrac{1}{2i}\cdot \sum\limits_{n=1}^{\infty}{\dfrac{(-1)^n n}{(2i)^n}({z-i})^{n-1}}=\dfrac{-1}{2i}\cdot \sum\limits_{n=1}^{\infty}{\dfrac{(-1)^n n}{(2i)^n}({z-i})^{n-3}}.$$ Then the principal part of Laurent expansion is (for $n=1,2$) $$\dfrac{-1}{2i}\left({\dfrac{-1}{2i}({z-i})^{-2}}+{\dfrac{2}{(2i)^2}({z-i})^{-1}}\right)$$