Find the Taylor series for $f(z)=\frac{i}{(z-i)(z-2i)}$ about $z_0=0$.

Question

Find the Taylor series for $f(z)=\frac{i}{(z-i)(z-2i)}$ about $z_0=0$ and the disk of convergence.

For the Taylor series I got $-\frac{i}{2}+\frac{z(2+i)}{4}-\frac{z(2i+3)}{2!}...$, but I'm not super confident in it. Can someone confirm or deny if it's correct?

To find this I simply used the $(_0)+\frac{′(_0)}{1!}(−0)+\frac{″(_0)}{2!}(−_0)^2+⋯$ expansion. I double checked with a derivative calculator so I know my values are right, I just wanted to double check that this is the right way to do it. I always thought it was, but that's not how my complex analysis class does it and I don't understand their way.

Answer 1

Note that, if $z\in\Bbb C\setminus\{i,2i\}$,\begin{align}\frac i{(z-i)(z-2i)}&=\frac1{i-z}-\frac1{2i-z}\\&=-\frac i{1+iz}+\frac12\frac i{1+iz/2}.\end{align}Therefore, if $|z|<1$,\begin{align}\frac i{(z-i)(z-2i)}&=-i\sum_{n=0}^\infty(-iz)^n+\frac i2\sum_{n=0}^\infty\left(-\frac{iz}2\right)^n\\&=\sum_{n=0}^\infty\left((-i)^{n+1}-\left(-\frac i2\right)^{n+1}\right)z^n\\&=\sum_{n=0}^\infty(-i)^{n+1}\left(1-\frac1{2^{n+1}}\right)z^n.\end{align}So, this last power series is the Taylor series that you're after.

Answer 2

Answer regarding the disc of convergence:

Theorem

Let $f:\Omega \to \Bbb C$ be holomorphic and let $z_0 \in \Omega$. Then, $f$ admits a power series representation centred at $z_0$. Moreover, this power series agrees with $f$ on the maximal open disc $D \subset \Omega$ centred at $z_0$.

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In this case, we have $\Omega = \Bbb C\setminus \{i, 2i\}$ and $z_0 = 0$.
Thus, the $D$ as in the theorem will be $D = \{z : |z| < 1\}$.
(Verify that $D$ is indeed contained $\Omega$ and that any larger disc would contain $i$.)

Thus, the power series (centred at $z_0 = 0$) $\sum a_nz^n$ that you derived will be equal to $f$ on $D$.

Now, here is the claim that we want to make:

Claim. $D$ is the disc of convergence of $\sum a_nz^n$.
Note that the above claim says precisely two things:

1. $\sum a_nz^n$ converges for every $z \in D$.
2. If we take any larger disc $D' \supsetneq D$, then there exists $z \in D'$ such that $\sum a_nz^n$ diverges.

Proof. The proof of 1. follows from the Theorem.
Let us focus on 2.

Let $D'$ be any (strictly) larger disc and let us assume that the power series converges on it.
Then, the function $$g(z) = \sum a_nz^n$$ is well-defined on $D'$. Moreover, it is continuous on $D'$. (A standard fact about power series.)

Also, we know that $f$ and $g$ agree on $D$. Moreover, $i \in D'$.
(In particular, $g(i)$ is defined and is a (finite) complex number.)

Now, let us take any sequence $(z_n)$ in $D$ such that $z_n \to i$. Since $f$ and $g$ agree on $D$, we get that $$f(z_n) = g(z_n)$$ for all $n \in \Bbb N$. However, since $f$ and $g$ are continuous, we see that $$|f(z_n)| \to \infty; \quad |g(z_n)| \to |g(i)| < \infty,$$ a contradiction! (The fact that $|f(z_n)| \to \infty$ follows from the function definition of $f$.)