Complex Taylor Series expansion - basic examples with radius of convergence

Question

Find radius of convergence and Taylor Series $\sum_{n=0}^{\infty}a_n(z-p)^n$ of $f$ where

1. $f(z) = \frac{1}{1-iz}$ with $p=0$
2. $f(z) = \frac{1}{1-z}$ with $p = i$
3. $f(z) = \frac{1}{1-z}$ with $p = 2$

Now, I feel like I can't just use

$$ \frac{1}{1-z} = \sum_{n=0}^{\infty}z^n $$

and "plug in" $z = iz$ or move $p$ around and say that for example $\frac{1}{1-z}$ around $p = 2$ is just $\sum_{n=0}^{\infty}(z-2)^n$ with radius of convergence 1. What is the correct way to do those examples? Any hints appreciated.

Answer 1

For $p=0$, that's precisely what you should do.

For other points $p$, you do need to "move $p$ around", so to speak, but not in the way you did it in your post. Just substituting $(z-2)$ for $z$ into the power series gives you a series for a different function: $$\sum_{n=0}^{\infty}z^n=\frac{1}{1-z} \implies \sum_{n=0}^{\infty}(z-2)^n=\frac{1}{1-(z-2)}=\frac{1}{3-z}\ne\frac{1}{1-z}.$$

What you should do instead is rewrite the original function first so that $(z-2)$ appears in it before expanding it into a Taylor series: $$\begin{aligned} \frac{1}{1-z}&=\frac{1}{1-z+2-2}=\frac{1}{1-(z-2)-2}=\frac{1}{-1-(z-2)}= \\ &=-\frac{1}{1+(z-2)}=-\frac{1}{1-(-(z-2))}= \\ &=-\sum_{n=0}^{\infty}(-(z-2))^n=-\sum_{n=0}^{\infty}(-1)^n(z-2)^n=\sum_{n=0}^{\infty}(-1)^{n+1}(z-2)^n. \end{aligned}$$ And the radius of convergence is determined by the fact that we substituted $(z-2)$ for $z$, so $|z-2|<1$.

Similarly, you can handle the remaining example.

Answer 2

You can replace $z$ with $iz$ in $\frac1{1-z}=\sum_{n=0}^\infty z^n$ to get the answer to the first question.

Concerning the second one, you can do\begin{align}\frac1{1-z}&=\frac1{1-i-(z-i)}\\&=\frac1{1-i}\cdot\frac1{1-\frac{z-i}{1-i}}\\&=\frac1{1-i}\sum_{n=0}^\infty\left(\frac{z-i}{1-i}\right)^n\text{ (when $\left\lvert\frac{z-i}{1-i}\right\rvert<1$)}\\&=\sum_{n=0}^\infty\frac{(z-i)^n}{(1-i)^{n+1}}\end{align}and the radius of convergence is $\sqrt2\bigl(=\lvert1-i\rvert\bigr)$.

Can you do the third one now?