Find a power series expansion for $\frac{1}{z}$ around $z = 1 + i.$
My Solution
For any complex $\alpha$,$\frac{1}{z}=\frac{1}{\alpha+z-\alpha}=\frac{1}{\alpha[1+\frac{z-\alpha}{\alpha}]}$ $=\frac{1}{\alpha}[1-\frac{z-\alpha}{\alpha}$$+\frac{(z-\alpha)^{2}}{\alpha^{2}}]-+...]$
$\displaystyle\sum_{k=0}^\infty \frac{(-1)^{k}(z-\alpha)^{k}}{\alpha^{k+1}}$ and then $\alpha=1+i$
Does anyone can help me improve it?