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?
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 $z=i$. $$\frac{1}{1-z} = \frac{1}{1-(z-i) - i} = \frac{1}{1-i}\frac{1}{1 - \left(\frac{z-i}{1-i} \right)} = \frac{1}{1-i}\sum\limits_{n=0}^{\infty}\left(\frac{z-i}{1-i}\right)^n$$