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Determine the Laurent series of the function $f(z)=\frac{1}{z^2+1}$ on the set $A=\{z|\:0<|z-i|<2\}$.

I know I should use the expansion $\frac{1}{z-1}=\sum_\limits{n=0}^{\infty}z^n$ or $\frac{1}{z+1}=\sum_\limits{n=0}^{\infty}(-1)^n z^n$.

The solution to the problem provided by the book is $\frac{-i}{2(z-i)}+\frac{1}{4}\sum_\limits{n=0}^{\infty}(\frac{-1}{2i})^n (z-i)^n$.

I have solve lots of different problems but the ring $A$ is always centred around $|z|$ not $|z-i|$.

Question:

How should I solve the exercise?

Thanks in advance!

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  • 3
    $\begingroup$ On way is to write the partial fraction decomposition $\frac{1}{z^2+1}=\frac{-i/2}{z-i}+\frac{i/2}{z+i}$. Observe that the first sumand is already a power of $z-i$. So, you only need to write the other one as powers of $z-i$. You can do that by subtracting and adding $i$ and using your formulas. $\frac{i/2}{z+i}=\frac{i/2}{(z-i)+2i}=\frac{i/2}{2i}\frac{1}{(z-i)/2i+1}$. Now use your second formula with $(z-i)/2i$ playing the role of $z$. $\endgroup$ – user647486 Apr 20 at 14:56
  • $\begingroup$ The Laurent series of all rational functions can be computed using this algorithm. $\endgroup$ – user647486 Apr 20 at 14:58
  • $\begingroup$ @user647486 Thanks a lot. I have already finished the exercise. $\endgroup$ – Pedro Gomes Apr 20 at 15:54
  • $\begingroup$ Write it here for the next fellow and to get it checked. $\endgroup$ – user647486 Apr 20 at 15:56
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You have$$\frac1{z^2+1}=\frac1{(z+i)(z-i)}.$$But\begin{align}\frac1{z+i}&=\frac1{2i+(z-i)}\\&=\frac1{2i}\frac1{1+\frac{z-i}{2i}}\\&=\frac1{2i}\sum_{n=0}^\infty(-1)^n\left(\frac{z-i}{2i}\right)^n\\&=\sum_{n=0}^\infty\frac{(-1)^n}{(2i)^{n+1}}(z-i)^n\end{align}and therefore\begin{align}\frac1{z^2+1}&=\frac1{(z+i)(z-i)}\\&=\sum_{n=0}^\infty\frac{(-1)^n}{(2i)^{n+1}}(z-i)^{n-1}\\&=\sum_{n=-1}^\infty\frac{(-1)^{n+1}}{(2i)^{n+2}}(z-i)^n.\end{align}

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