# Laurent Series, region of convergence

I want to find the laurent series for $$f(z) = \frac{z}{z^2 - (1+i)z +i}$$ in powers of $z-1$ and find the region of convergence. I am not quite sure how to do this. I know that $$f(z) = \frac{z}{(z-1)(z-i) }$$ but I do not know where to go from here. Any help would be great! Thanks!

• I have not, I am unsure of how to find the series representation for this – Bob Math Dec 11 '14 at 0:45
• Use partial fractions decompositions of $f(z)$. – dustin Dec 11 '14 at 0:46
• So, you end up getting ((z-1) + 1) /( z(z-1) - i(z-1) ) = ( z-1 / z(z-1) - i(z-1) ) + 1/ (z(z-1) - i(z-1) ) – Bob Math Dec 11 '14 at 0:48
• Unless I made a mistake, I got $\frac{1}{i+1}\frac{1}{z-i}-\frac{i}{i+1}\frac{1}{z-1}$ dont distribute the constants. – dustin Dec 11 '14 at 0:51
• That is correct I just redid it – Bob Math Dec 11 '14 at 0:56

First rewrite $f$: $$f(z) = \frac{z}{z^2 - (1+i)z +i} = \frac{z}{(z-1)(z-i)} = \left(\frac{A}{z-1} + \frac{B}{z-i}\right) = \frac{1}{1-i}\left(\frac{1}{z-1} - \frac{i}{z-i}\right).$$ Since you want the expansion around $1$, let's write everything in terms of $(z-1)$: \begin{align} f(z) &=\frac{1}{1-i}\left(\frac{1}{z-1} - \frac{i}{(z-1)+(1-i)}\right)\\ \end{align} Note that $f$ has two singularities, one at $1$ and the other at $i$, so if you want the Laurent expansion around $1$ it has to be done for two different cases: $|z-1|<\sqrt{2}$ and $|z-1|>\sqrt{2}$. Then, for $|z-1|<\sqrt{2}$ we have that $|z-1|<|1-i|$, so: \begin{align} f(z) & = \frac{1}{1-i}\left(\frac{1}{z-1} - \frac{i}{1-i}\frac{1}{1 + \frac{z-1}{1-i}}\right)\\ &= \frac{1}{1-i}\left(\frac{1}{z-1} - \frac{i}{1-i}\sum_{k=0}^\infty (-1)^k\left(\frac{z-1}{1-i}\right)^k\right). \end{align} Analogously if $|z-1|>|1-i|=\sqrt{2}$: \begin{align} f(z) & = \frac{1}{1-i}\left(\frac{1}{z-1} - \frac{i}{z-1}\frac{1}{1 + \frac{1-i}{z-1}}\right)\\ &= \frac{1}{1-i}\left(\frac{1}{z-1} - \frac{i}{z-1}\sum_{k=0}^\infty (-1)^k\left(\frac{1-i}{z-1}\right)^k\right). \end{align}