So what I did was using partial fractions, I broke $f(z)$ down to \begin{align*}f(z)= \frac{1}{(z-i)(z-2)} = \frac{\frac{1}{i-2}}{z-i} + \frac{\frac{1}{2-i}}{z-2} \end{align*} and then got Taylor series expansions for each. Then I got a scarily messy equation \begin{align*}f(z)&=\frac{1}{i-2} (-1)^n (-i)^{-(n+1)} z^n-\frac{1}{i-2} (-1)^n (-2)^{-(n+1)} z^n\end{align*} And then it seems impossible for me to calculate $R$ using $\limsup$. Can anyone help me on this? Thanks
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Let's do some manipulations (I'll leave the constants aside for the rest of the post), $$ \frac{1}{z-i} = \frac{1}{-i}\frac{1}{1-\frac{z}{i}}. $$ Then, remember that for every $x\in\mathbb{C}$ such that $|x|<1$, $$ \frac{1}{1-x} = \sum_{k=0}^\infty x^k. $$ So, whenever $|z|<|i|=1$, we have that: $$ \frac{1}{z-i} = i \frac{1}{1-\frac{z}{i}} = i\sum_{k=0}^\infty \frac{z^k}{i^k}. $$ This is the Taylor series of the fraction around $0$ and as you can see, the radius of convergence is $1$.
Can you repeat this process for the other fraction and conclude?
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$\begingroup$ Thank you. I think I just needed to know how you manipulated the first equation. $\endgroup$ Nov 7, 2014 at 13:40
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$\begingroup$ You're just trying to get something of the form $1-x$ in the denominator of the fraction where $x=z/k$ with $k$ constant. The radius of convergence is precisely $|k|$. Observe that if you wanted the principal part of the Laurent series you'd have to do the same, just with $x=k/z$. $\endgroup$– hjhjhj57Nov 7, 2014 at 16:04