Develop $f(z) = (1- \frac{1}{z} )^3$ to Laurent's series around $z_{0} = i$.

Question

Develop $f(z) = (1- \frac{1}{z} )^3$ to Laurent's series around $z_{0} = i$.

I was thinking to cube it first and than observe each member of that sum of four elements and find Laurent's series for it. Is that good thinking?

Answer 1

Just do Taylor around $\;z_0=i\;$ :

$$\begin{cases}f(i)=\left(1-\frac1i\right)^3=(1+i)^3=2i(1+i)=-2+2i\\ f'(z)=\frac1{z^2}\implies f'(i)=\frac1{i^2}=-1\\ f''(z)=\frac{-2}{z^3}\implies f''(i)=-\frac2{i^3}=-2i\\ f'''(z)=\frac6{z^4}\implies f'''(i)=6\\ \ldots\\ f^{(n)}(z)=\frac{(-1)^{n+1}\,n!}{z^{n+1}}\implies f^{(n)}(i)=\frac{(-1)^{n+1}n!}{i^{n+1}}\\ \ldots\end{cases}$$

Thus, the series you're looking for is

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