Calculation of the Laurent Series

Question

How can one calculate the Laurent Series of: $$f(z)=\frac{1}{z(z-i)^2},\quad 0<|z-i|<1$$ So far i have calculated that $$f(z)=\frac{1}{z-i}-\frac{1}{(z-i)^2}-\frac{1}{z}$$

Answer 1

Let us start with $\frac1z$. We have\begin{align}\frac1z&=\frac1{i+z-i}\\&=\frac{-i}{1-i(z-i)}\\&=-i\sum_{n=0}^\infty\bigl(i(z-i)\bigr)^n\text{ (since $|z-i|<1$)}\\&=\sum_{n=0}^\infty-i^{n+1}(z-i)^n,\end{align}Therefore$$\frac1{z(z-i)^2}=\sum_{n=0}^\infty-i^{n+1}(z-i)^{n-2}=\sum_{n=-2}^\infty-i^{n+3}(z-i)^n.$$

Answer 2

The correct answer appears elsewhere. However, you have made a significant error in your $\text{"}$calculated that $$f(z)=\frac{1}{z-i}-\frac{1}{(z-i)^2}-\frac{1}{z} \text{."}$$

This is a job for partial fractions. $$ \frac{1}{z(z-\mathrm{i})^2} = \frac{A}{z} + \frac{B}{z-\mathrm{i}} + \frac{C}{(z-\mathrm{i})^2} \text{.} $$ Clear denominators and collect by power of $z$. \begin{align*} 1 &= \frac{A}{z}z(z-\mathrm{i})^2 + \frac{B}{z-\mathrm{i}}z(z-\mathrm{i})^2 + \frac{C}{(z-\mathrm{i})^2}z(z-\mathrm{i})^2 \\ &= A(z-\mathrm{i})^2 + Bz(z-\mathrm{i}) + Cz \\ &= (A+B)z^2 + (-2\mathrm{i}A - \mathrm{i}B + C)z - A \text{.} \end{align*} For two (real or complex) polynomials to be equal, they must have equal coefficients, so we get the system $$ \begin{cases} 0 &= A+B \\0 &= -2\mathrm{i}A - \mathrm{i}B + C \\ 1 &= -A \end{cases} \text{.} $$ We immediately see $A = -1$ and $B = 1$ and this gives $C = -\mathrm{i}$. Therefore, $$ f(z) = \frac{-1}{z} + \frac{1}{z-\mathrm{i}} + \frac{-\mathrm{i}}{(z-\mathrm{i})^2} \text{.} $$