Let $f$ be the function :

$$f(z) = \bigg(\frac{z}{z+1}\bigg)^2 = \bigg(1-\frac{1}{1+z}\bigg)^2 = 1 -2\frac{1}{1+z} + \frac{1}{(1+z)^2}$$

Find the Taylor expansion of the function $f$ around $i$ and also the convergence radius.


From the Cauchy-Taylor theorem :

$$f(z) = \sum_{n=0}^{\infty} a_n(z-i)^n = \sum_{n=0}^{\infty} \frac{f^{(n)}(i)}{n!}(z-i)^n,|z-i|<R $$

We have to calculate the $a_n = \frac{f^{(n)}(i)}{n!}, \forall n\in \mathbb N$ and the radius of convergence $R$.

Now, in similar exercises, all fraction function could be manipulated to give you such a form around the expansion point, that could be used to be set as $w$, in order to use one of the known geometric series :

$$\frac{1}{1-w} = \sum_{n=0}^{\infty}w^n ,|w|<1$$ $$\frac{1}{1+w} = \sum_{n=0}^{\infty}(-1)^nw^n,|w|<1$$

And after having such an expression, by differentiating the geometric series, you'd be getting the final expression of the Taylor Expansion, since the function has one and only series expansion. Finally, to get the convergence radiues, you would just solve the expression $|w|<1$, where $w$ would be a fraction-function of $z-i$.

My problem with the particular example is that I cannot seem to find a way (or my head is stuck) to make such an expression, thus I cannot continue with the steps I've mentioned.

I'd appreciate a thorough help/solution or any advice and tips on such problems, because it's a new subject I'm working on, on our Complex Analysis Semester.

Thanks for your time!

  • 2
    $\begingroup$ Hint: $$\frac1{1+z}=\frac1{1+i+z-i}=\frac1{1+i}\frac1{1+\frac{z-i}{1+i}}$$ $\endgroup$ May 28, 2017 at 11:30
  • 1
    $\begingroup$ One approach would be via $\dfrac{1}{(1+z)^2}=\dfrac{1}{((1+i)+(z-i))^2}=\dfrac{1}{(1+i)^2}\dfrac{1}{(1-\frac{z-i}{-1-i})^2}=$ $=c\dfrac{1}{(1-u)^2}=c\dfrac{d}{du}\dfrac{1}{1-u}=c\dfrac{d}{du}\sum_{n=0}^\infty u^n$ where you end up with a double sum. $\endgroup$
    – Nikolaj-K
    May 28, 2017 at 11:31

2 Answers 2


A good strategy is to set $z=w+i$ and find the Taylor expansion at $0$. The function becomes $$ g(w)=f(w+i)=\frac{(w+i)^2}{(w+i+1)^2} $$ and it is clear you need the Taylor expansion of $$ \frac{1}{(w+i+1)^2}= \frac{1}{(1+i)^2}\frac{1}{\bigl(1+\frac{w}{1+i}\bigr)^2} $$ Set $v=-w/(1+i)$: you now need the expansion of $$ \frac{1}{(1-v)^2} $$ which is known: integrate and differentiate. Then substitute, multiply by $w^2+2iw-1$ and change back $w=z-i$.


Derivatives of $f(z)$ are not so hard to find, after all: $$ f^{(n)}(z)=2(-1)^{n+1}{n!\over(1+z)^{n+1}}+(-1)^{n+2}{(n+1)!\over(1+z)^{n+2}} ,\quad n\ge1. $$ Substituting here $z=i$ and taking into account that $1+i=\sqrt2e^{i\pi/4}$, we can find: $$ a_n={f^{(n)}(i)\over n!}=\left({e^{3i\pi/4}\over\sqrt2}\right)^{n+1} \left(2+(n+1){e^{3i\pi/4}\over\sqrt2}\right), \quad n\ge1, $$ and of course $a_0=f(i)={i\over2}$.

The radius of convergence can be found from $R=\lim\limits_{n\to+\infty}\left|{a_n\over a_{n+1}}\right|=\sqrt2$, which is obvious, as $f(z)$ has a pole at $z=-1$.


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