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Find a power series expansion for $\frac{1}{z}$ around $z = 1 + i.$

My Solution

For any complex $\alpha$,$\frac{1}{z}=\frac{1}{\alpha+z-\alpha}=\frac{1}{\alpha[1+\frac{z-\alpha}{\alpha}]}$ $=\frac{1}{\alpha}[1-\frac{z-\alpha}{\alpha}$$+\frac{(z-\alpha)^{2}}{\alpha^{2}}]-+...]$

$\displaystyle\sum_{k=0}^\infty \frac{(-1)^{k}(z-\alpha)^{k}}{\alpha^{k+1}}$ and then $\alpha=1+i$

Does anyone can help me improve it?

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    @$A$ndréNicolas Done. – 2012-05-01

1 Answers 1

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When one defines a power series, it is also necessary to define what the domain of convergence is. It's basically similar as to when we define a function: we have to define a domain. You want to express the function

$f(z)=\frac 1 z $

as a power series around $z=i+1$. You cleverly now that the function

$ g(z)=\frac{1}{z-a}$

can be expanded as

$ g(z)=\frac{1}{z-a} =-\frac 1 a\frac{1}{1-z/a} =-\frac 1 a \sum_{n=0}^\infty \left(\frac z a \right)^n$

But when is this representation legitimate? If we plug any $z$ such that $|z|>a$, we'll find ourself with a divergent series. So, what we should really write is

$ g(z)=-\frac 1 a \sum_{n=0}^\infty \left(\frac z a \right)^n\text{ ; }\color{red}{ |z|

Note that it can't be the case that $a=0$.

Moving on to your problem. We have that,

$f(z)=\frac 1 z =\frac 1 {z-a+a}=-\frac 1 a \frac 1 {1-\frac {z+a}{a}}$ so again we write

$f(z)=-\frac 1 a \frac 1 {1-\frac {z+a}{a}}=-\frac 1 a \sum_{n=0}^\infty \left(\frac {z+a}{a} \right)^n$

Note that again it cant be the case that $a=0$. And similarily, we need that

$|z+a|<|a|$ for the series to converge. So our domain of convergence will be

$\Bbb D = \{ z \in \Bbb C : |z+a|<|a|\}$

In this case we want to expand around $z=i+1$, so we choose $a=-(i+1)$, which gives

$f(z)=\frac 1 {i+1} \sum_{n=0}^\infty (-1)^n \left(\frac {z-(i+1)}{i+1} \right)^n$

$\frac 1 {i+1}=\frac {1-i}{2}$

so we can write this as

$f(z)=\frac {1-i}{2} \sum_{n=0}^\infty (-1)^n \left(\frac {1-i}{2}\right)^n \left( z-(i+1) \right)^n$

As a final step, we find what $\Bbb D$ should be.

$|z-(i+1)|<|i+1|$

$|z-(i+1)|< \sqrt 2$

So the series can be used in $\Bbb D$, a disk of radius $\sqrt 2$ and center at $z=i+1$.