Find the Laurent series of $\frac{1}{1+x^2}$ centered around $i$

Question

I am having trouble with finding the Laurent series of the following function, centered around $i$:

$$f(x) = \frac{1}{1+x^2} .$$

I tried transforming this into a form that would be appropriate for a geometric series, but was not able to.

Answer 1

Hint Use the Method of Partial Fractions to find the coefficients in the decomposition $$\frac{1}{1 + x^2} = \frac{A}{1 + i x} + \frac{B}{1 - i x} .$$ NB that since the left-hand side is real, so is the right-hand side, and hence $B = \bar{A}$.

Answer 2

Let $z = x - i$. Then $$ f = \frac{1}{1+x^2} = \frac{1}{1 + (i + z)^2} = \frac{1}{z^2 + 2iz} = \frac{1}{z} \frac{1}{z + 2i} $$ We are trying to find its series expansion at $z = 0$. Let's temporarily discard the first factor. Since $1/(z+2i)$ is holomorphic, $$\begin{align*} \frac{1}{z + 2i} &= \sum_{j=0}^\infty a_j z^j \\ 1 &= \sum_{j=1}^\infty a_{j-1} z^j + 2i \sum_{j=0}^\infty a_j z^j \\ &= 2i a_0 + \sum_{j=1}^\infty (a_{j-1} + 2i a_j) z^j \end{align*}$$ Substituting $z = 0$ yields $a_0 = -i/2$. Since all other terms are $0$, we can establish a recurrence relation $$ a_j = \frac{i}{2}a_{j-1} = -\left( \frac{i}{2} \right)^{j+1} $$ Therefore, $$ \frac{1}{z + 2i} = \sum_{j=0}^\infty -\left( \frac{i}{2} \right)^{j+1} z^j $$ Don't forget to reintroduced the $z^{-1}$ term and substitute $z = x - i$.