# Find the Taylor series of $\frac{1+z}{1-z}$ at $z_{0}=i$

I'm given the following explanation:

Let $$\dfrac{1+z}{1-z} = \biggl(\dfrac{1+i}{1-i}+\dfrac{z-i}{1-i}\biggr)\biggl(1-\dfrac{z-1}{1-i}\biggr)^{-1}$$ =

= $$\dfrac{1+i}{1-i}\displaystyle\sum_{j=0}^{\infty}\biggl(\dfrac{z-i}{1-i}\biggr)^{j} + \displaystyle\sum_{j=0}^{\infty} \biggl(\dfrac{z-i}{1-i}\biggr)^{j+1}$$

I've been staring at this problem for quite some time. Do I first apply the product rule and plug in $$z_{0}=i$$ for each consecutive derivative?

## 1 Answer

This is pretty generic problem and a standard technique is to bring the function into a geometric series. $$\frac{1+z}{1-z} = \frac{(1-z)(-1) + 2}{1-z} = -1 + \frac{2}{1-z}.$$ Now, you only expand $$1/1-z$$ around $$z=i$$. $$\frac{1}{1-z} = \frac{1}{1-(z-i) - i} = \frac{1}{1-i}\frac{1}{1 - \left(\frac{z-i}{1-i} \right)} = \frac{1}{1-i}\sum\limits_{n=0}^{\infty}\left(\frac{z-i}{1-i}\right)^n$$

• Does the -1 have any use? – K.M Apr 14 at 9:07
• It is not only -1. You get additional constant term from n=0. Otherwise left and right hand side do not match. – WoofDoggy Apr 14 at 14:06