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 1


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

  • $\begingroup$ Does the -1 have any use? $\endgroup$
    – Skm
    Commented Apr 14, 2019 at 9:07
  • $\begingroup$ It is not only -1. You get additional constant term from n=0. Otherwise left and right hand side do not match. $\endgroup$
    – WoofDoggy
    Commented Apr 14, 2019 at 14:06

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