Taylor Series expansion of $f(z)=(1+z)/(1-z)$

Question

I am trying to determine the Taylor Series expansion of $f(z)=\frac{(1+z)}{(1-z)}$ centered at $z_0=i$ by trying to rewrite it in terms of a geometric series. However, I wasn't able to proceed using this idea. What other methods could be applied in order to solve this problem?

Answer 1

Hint: $f(z)=\frac{1+z}{1-z}=-1+\frac{2}{1-z}=-1+\frac{2}{(1-i)-(z-i)}=-1+\frac{2}{1-i}\frac{1}{1-(\frac{z-i}{1-i})}$ .

Now use this to do what you were trying to do before.

Answer 2

Hint: for the series of $f(z)$ centred at $z=c$, it's convenient to substitute $z = c+t$ so that you're taking the series of $f(c+t)$ centred at $t=0$. Also notice that $$ \frac{1+c+t}{1-(c+t)} = -1 + \frac{2}{1-(c+t)} = -1 + \frac{2}{1-c}\left(\frac{1}{1-t/(1-c)}\right)$$

Answer 3

Using Robert Israel's hint, you should end with $$f(z)=\frac{(1+z)}{(1-z)}\implies g(t)=\frac{(1+i+t)}{(1-i-t)}$$ and then $$g(t)=i+\sum_{n=1}^\infty \frac{(1+i)^{n+1}}{ 2^{n}}\, t^n$$ which could be transformed using $$(1+i)=\sqrt 2\,e^{i \frac \pi 4}=\sqrt{2} \left(\cos \left(\frac{\pi }{4}\right)+i \sin \left(\frac{\pi }{4}\right)\right)$$ and de Moivre formula to get $$g(t)=i+\sqrt 2\sum_{n=1}^\infty 2^{-\frac{n}{2}}\left(\cos \left(\frac{(n+1)\pi}{4} \right)+i \sin \left(\frac{(n+1)\pi}{4} \right)\right)t^n$$