Laurent Series computation

Question

Need to compute the Laurent Series:

$$f(z) = \frac{1}{1+z^2} \quad on \quad 0<|z-i|<2$$

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$$f(z) = \frac{1}{2i(z-i)} + \frac{i}{2(z+i)}$$

$$\frac{1}{2i(z-i)} = \frac{1}{2zi}\sum_{k=0}^\infty \left(\frac{i}{z}\right)^k$$

$$\frac{i}{2(z+i)} = \frac{i}{2z}\sum_{k=0}^\infty \left(\frac{-i}{z}\right)^k$$

$$f(z)=\frac{i}{2z}\sum_{k=0}^\infty \frac{(-i)^k-(i)^k}{z^k}$$

Is this correct? I cannot figure out a way to verify the answer.

Answer 1

$$f(z) = \frac{1}{(z-i)(z+i)}$$

$$f(z) = \frac{1}{(z-i)}\cdot \frac{1}{(z+i)}$$

$$f(z) = \frac{1}{(z-i)}\cdot \frac{1}{2i-(i-z)}$$

$$f(z) = \frac{1}{(z-i)}\cdot \frac{1}{2i}\cdot\frac{1}{1-\frac{(i-z)}{2i}} $$

Clearly $\displaystyle \frac{1}{1-\frac{(i-z)}{2i}}=\sum_{n=0}^{\infty}\frac{(i-z)^n}{(2i)^n}$ whenever $\left|\frac{(i-z)}{2i} \right|<1 \Rightarrow 0<|z-i|<2$

$$f(z) = \frac{1}{(z-i)}\cdot \frac{1}{2i}\cdot\sum_{n=0}^{\infty}\frac{(i-z)^n}{(2i)^n}= \sum_{n=0}^{\infty}\frac{(-1)^n(z-i)^{n-1}}{(2i)^{n+1}}$$

$$f(z) =\sum_{n=0}^{\infty}\frac{(-1)^n(z-i)^{n-1}}{(2i)^{n+1}}$$

Answer 2

As I noted in the comments, for the given disk we are looking for a series of the form $$\sum_{k=-\infty}^{\infty} c_k (z-i)^k$$ You correctly calculated that $f(z) = \frac{1}{2i(z-i)} + \frac{i}{2(z+i)}$. Now you should note that the first term is $\frac{1}{2i}(z-i)^{-1}$ which is of the desired form so now we only need a series for the second term.

For convenience of notation let $w = z - i$. Then $$\frac{i}{2(z+i)} = \frac{i}{2}\frac{1}{w + 2i} = \frac{1}{4}\frac{1}{\frac{w}{2i} + 1}$$ Now note that $|\frac{w}{2i}|<1$ so this is the sum of the geometric series $$\frac{1}{\frac{w}{2i} + 1} =\sum_{k=0}^{\infty} \bigg (\frac{-w}{2i} \bigg)^k $$ This can all be put back together to get the correct Laurent series on the desired annulus. $$f(z) = \frac{1}{2i}(z-i)^{-1} + \frac{1}{4}\sum_{k=0}^{\infty} \bigg (\frac{-1}{2i} \bigg)^k(z-i)^k = \sum_{k=-1}^{\infty} \frac{1}{4} \bigg (\frac{-1}{2i} \bigg)^k(z-i)^k $$