Laurent series expansion of $f(z)=\frac {2(1-i)z} {z^2-2(1+i)z+4i}$ in $|z|>2$

Question

Can anybody guide me how to find Laurent series expansion $f(z)=\frac {2(1-i)z} {z^2-2(1+i)z+4i}$ in $|z|>2$, I tried to complete the square in the denominator, but couldn't finish

Answer 1

Note that $(1+i)^2=2+2i-2=2i,$ so we can write $$z^2-2(1+i)+4i=\bigl(z-(1+i)\bigr)^2+2i=\bigl(z-(1+i)\bigr)^2+(1+i)^2.$$ Using the factoring identity $$a^2+b^2=(a+ib)(a-ib)$$ for complex numbers $a$ and $b$, and noting that $i(1+i)=-1+i$, we can then write $$z^2-2(1+i)+4i=(z-2)(z-2i).$$ Now, let's use partial fractions to write $$\frac{2(1+i)z}{z^2-2(1+i)z+4i}=\frac{A}{z-2}+\frac{B}{z-2i},$$ so that $$2(1+i)z=A(z-2i)+B(z-2).\tag{$\star$}$$ Putting $z=2$ in $(\star)$ gives us $$A=\frac{2(1+i)}{1-i}=\frac{2(1+i)^2}{2}=2i,$$ and so $B=2$ by back-substitution into $(\star)$. Thus, we can write $$\frac{2(1+i)z}{z^2-2(1+i)z+4i}=\frac{2i}{z-2}+\frac{2}{z-2i},$$ and for $z\neq 0,$ this can be rewritten as $$\frac{2(1+i)z}{z^2-2(1+i)z+4i}=z\left(\cfrac{2i}{1-\frac2z}+\cfrac{2}{1-\frac{2i}z}\right)=2iz\cdot\cfrac1{1-\frac2z}+2z\cdot\cfrac1{1-\frac{2i}z}.\tag{#}$$ Now, when $|z|>2$ then both $\left|\frac2z\right|<1$ and $\left|\frac{2i}z\right|<1$, so we can rewrite the fractional expressions at the far right of $(\#)$ as geometric series. Once that's done, getting to the desired Laurent series will be fairly simple.