Taylor series in complex plane

Question

• What is the Taylor series and radius of convergence for $\frac{1+z}{1-z}$ at $z_0=i$.

There was a hint provided to convert $z$ to $i+(z-i)$ and use the geometric series.

I got $$\frac{1+i+z-i}{1-i-(z-i)}$$ which looks like a geometric series for $(z-i)$ but I could not continue.

Answer 1

First of all, the radius of convergence $R$ of the Taylor series of $f$ at $z_0$ can be found independently as the distance from $z_0$ to the first pole of $f$ you meet when expanding (inflating) a disk centered in $z_0$. This first (besides unique) pole is here $z_1=1$.

As $z_0=i$, we have

$$R=|z_0-z_1|=|i-1|=\sqrt{2}.$$

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Now, the Taylor series. Transform your expression (as you started to do) by dividing its numerator and denominator by $1-i$ like this :

$$\frac{1+z}{1-z}=\frac{1+i+z-i}{1-i-(z-i)}=\frac{\frac{1+i}{1-i}-\frac{z-i}{1-i}}{1-\frac{z-i}{1-i}}=\frac{i+Z}{1-Z}=(i+Z)\frac{1}{1-Z}\tag{1}$$

where $$Z:=\frac{z-i}{1-i}.$$

The essential condition

$$|Z|<1 \iff |z-i|<|1-i|=\sqrt{2}=R$$

(the radius of convergence already found by another means...)

ensures that we can expand (1) into a convergent geometric series :

$$(i+Z)\sum_{n=0}^{\infty} Z^n$$

From here, it easy to deduce the Taylor series of $f$ at $z_0=i$ :

$$\sum_{n=0}^{\infty} a_n (z-i)^n$$

Answer 2

Just to kick the extra point on @Jean Marie's nice answer, the Taylor series is $f(z)=\sum_{n\ge0} a_n(z-i)^n$, where $a_n=(i+1)/(1-i)^n\,,n\ge1$ and $a_0=i$, with radius of convergence $\sqrt2$.