1

I am preparing for an exam and i am trying to solve an exercise about Laurent expansion of a function. The book has no so solutions or end results, so i can't control my way of solving the exercise. I would like to ask you if my idea is correct and if you have any remarks, i would be glad if you give me a hint where to make a correction.

We have to find a Laurent expansion of the function $f(z)=\frac{1}{1+z^{2}}$ with given $z_{0}$ with $Im(z_{0})<0$ and $r=\left | z_{0} + i\right |$ and $R=\left | z_{0} - i\right |$.

I start from $f(z)=\frac{1}{1+z^{2}}=\frac{1}{2i}\frac{1}{z-i}-\frac{1}{2i}\frac{1}{z+i}$

As we should give the expansion in $r<\left | z-z_{0} \right |<R$,

I start with $r<\left | z-z_{0} \right |$ or $\left | z_{0} +i\right |<\left | z-z_{0} \right |$ $\Leftrightarrow 1 >\frac{z_{0}+i}{z-z_{0}}$, because we need to build a convergent geometric series. By transforming the second product i get: $-\frac{1}{2i}\frac{1}{z+i}=\frac{1}{2i}\sum_{v<0} \left ( -\frac{z_{0}+i}{z-z_{0}} \right )^v$.

The same i do for $\left | z-z_{0} \right |<R \Leftrightarrow \left | z-z_{0} \right | <\left | z_{0}-i \right |\Leftrightarrow \frac{\left | z-z_{0} \right |}{\left | z_{0}-i \right |}<1$. I get: $\frac{1}{2i}\frac{1}{z-i}=\frac{1}{2i}\frac{1}{z-z_{0}}\frac{1}{1-\frac{i-z_{0}}{z-z_{0}}}=\frac{1}{2i}\sum_{v<0} \frac{(z-z_{0})^{v}}{(i-z_{0})^{(v+1)}}$.

What i have to do at the end is to calculate the sum of the two series.

Is my solution correct or you see some problems in it? Thank you very much in advance!

Git Gud
  • 31,356
Lullaby
  • 2,361

1 Answers1

4

It's almost correct. The principle is right, but you got your series slightly wrong. We have

$$\begin{align} \frac{1}{z+i} &= \frac{1}{(z-z_0) + (z_0+i)}\\ &= \frac{1}{z-z_0}\cdot \frac{1}{1 + \frac{z_0+i}{z-z_0}}\\ &= \frac{1}{z-z_0}\sum_{\nu = 0}^\infty \left(-\frac{z_0+i}{z-z_0}\right)^\nu\\ &= \sum_{\nu=0}^\infty (-1)^\nu\frac{(z_0+i)^\nu}{(z-z_0)^{\nu+1}} \end{align}$$

and

$$\begin{align} \frac{1}{z-i} &= \frac{1}{(z-z_0) + (z_0-i)}\\ &= \frac{1}{z_0-i}\cdot \frac{1}{1 + \frac{z-z_0}{z_0-i}}\\ &= \frac{1}{z_0-i}\sum_{\nu=0}^\infty \left(-\frac{z-z_0}{z_0-i}\right)^\nu\\ &= \sum_{\nu=0}^\infty (-1)^\nu \frac{(z-z_0)^\nu}{(z_0-i)^{\nu+1}} \end{align}$$

Daniel Fischer
  • 206,697