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I just ran into the following exercise:

Find and state the convergence properties of the Taylor series for the following: $$\frac{1+z}{1-z}$$ around $z_0=i$.

First of all, let

$$f(z)=\frac{1+z}{1-z}.$$

Then we have that

$$\begin{align} f^{(1)}(z)&=\frac{2}{(1-z)^2},\\ f^{(2)}(z)&=\frac{-4}{(1-z)^3},\\ f^{(3)}(z)&=\frac{12}{(1-z)^4},\cdots \end{align}$$

leading us to conclude that

$$\frac{d^j}{dz^j}\left[\frac{1+z}{1-z}\right]=2(-1)^{j+1}\left(1-z\right)^{-j-1}j!,$$

for $j\geqslant1$. Now, evaluating these at $z=i$ should give the Taylor series

$$i+\sum_{j=1}^{\infty}\frac{2}{(1-i)^{j+1}}(z-i)^j.$$

I have two questions: I could derive that same series without problems if the $(-1)^{j+1}$ term were not there, but in this case, where does it go? Also, how does the author determine that this holds whenever $|z-i|<\sqrt{2}$? I suppose that this has something to do with the fact that

$$\sum_{j=0}^{\infty}c^j$$

converges whenever $|c|<1$. Thanks in advance!

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22 minutes. $ $ –  Did Jul 5 '12 at 16:36
    
Josue, if you dont mind how did you know to put the $i+$ in $i+\sum_{j=1}^{\infty}\frac{2}{(1-i)^{j+1}}(z-i)^j$ also how did you get the $(z-i)^j$ from? –  Q.matin Mar 4 '13 at 0:03

3 Answers 3

up vote 2 down vote accepted

You shouldn't have that alternating sign term showing up. $$f''(z)=\frac{d}{dz}\bigl[2(1-z)^{-2}\bigr]=2\cdot-2(1-z)^{-3}\cdot\frac{d}{dz}[1-z]=-4(1-z)^{-3}\cdot-1=\frac{4}{(1-z)^3}.$$ Something similar will happen in all other cases, too.

It holds whenever $|z-i|<\sqrt{2}$ because the nearest (in fact, only) singularity is the pole at $z=1$. The distance from $i$ to $1$ is $\sqrt{2}$. The radius of convergence of a Taylor series is the distance from the center to the nearest point of nonanalyticity. You can also, of course, see this by rearranging this as a geometric series. Note that $\frac{2}{1-i}=1+i$, so we can rewrite as $$\frac{1+z}{1-z}=i+(1+i)\sum_{j=1}^\infty\left(\frac{z-i}{1-i}\right)^j=-1+(1+i)\sum_{j=0}^\infty\left(\frac{z-i}{1-i}\right)^j,$$ and the series converges for $$\left|\frac{z-i}{1-i}\right|<1,$$ or equivalently, for $$|z-i|<|1-i|=\sqrt{2}.$$

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You are right! I had forgotten to apply the chain rule thoroughly. Also, I see it now: the pole is at $1$, and $|1-i|=\sqrt{2}$. Thank you! –  Josué Molina Jul 5 '12 at 16:34

I prefer using geometric series tricks to taking derivatives. Your fraction is $$ \frac{(1+i)+ (z-i)}{(1 - i) - (z-i)}. $$ Using $(1+i)(1-i) = 2$, multiply numerator and denominator by $\frac{1+i}{2}$ to get $$ \frac{\text{stuff}}{1 - \frac{(1+i)(z-i)}{2}} $$ and so the radius of convergence is $|\frac{2}{1+i}| = \sqrt{2}$ (and you get the explicit formula in the usual way if you're less lazy than I am with the numerator).

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$$ \frac{1+i+u}{1-i-u}=-1+\frac2{1-i}\frac1{1-\frac{u}{1-i}}=-1+\sum\limits_{n=0}^{+\infty}\frac2{(1-i)^{n+1}}u^n,\qquad |u|\lt|1-i|=\sqrt2. $$

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