Expand $\frac{1}{(z-1)(z-2i)}$ so that it converges when $|z|<1,1<|z|<2,2<|z|$

Question

I've tried looking at this using partial fractions: $\frac{1}{(z-1)(1-2i)}-\frac{1}{(z-2i)(1-2i)}$. However this just seems to complicate matters (or I've done the PFD incorrectly).

Answer 1

Your PFD should look like $\frac{1}{(z-1)(z-2i)}=\frac{A}{z-1}+\frac{B}{z-2i}$. But it's a fruitful approach in general. Just remember a very famous series.

Answer 2

Like you said, $$f(z)=\frac{1}{(z-1)(1-2i)}-\frac{1}{(z-2i)(1-2i)}= \frac{1}{-1+2i}\cdot\frac{1}{1-z} + \frac{1}{1-2i}\cdot \frac{1}{2i-z}$$

Now, turn each part into a Laurent series by using the geometric sum formula:

$$ |x|<1 \Leftrightarrow \frac{1}{1-x} = \sum_{n=0}^\infty x^n $$

If you want to write a Laurent series for $\frac{1}{1-z}$ that converges at $|z|<1$, easy peasy.

But if you want to write the series to converge at $|z|>1$, there is a trick: $$ \frac{1}{1-z} = \frac{1}{z} \cdot \frac{1}{\frac{1}{z} - 1 } = -\frac{1}{z} \cdot \frac{1}{1-\frac{1}{z} } = -\frac{1}{z} \cdot \sum_{n=0}^\infty \left ( \frac{1}{z} \right ) ^n = - \sum_{n=0}^\infty z^{-(n+1)}=$$ $$- \sum_{n=-\infty}^{-1} z^n $$

And the sum formula is true only for $|\frac{1}{z}|<1 \Rightarrow |z|>1 $. Mission accomplished.

Now do the same for the second part to get a series for $|z|>|2i|=2$ and for $|z|<|2i|=2$ , and for each area in $\mathbb{C}$ add the two series that converge in that area.

I hope this helped!