1
$\begingroup$

I've been trying to understand how to solve this Laurent series with partial fractions and substitutions, but I can't seem to get it right;

$$0<|z-2i|<R$$ $$f(z) = \frac{1}{z^2+4} = \frac{1}{(z-2i)(z+2i)}$$

I'm having a hard time understanding how to work with the zero, and also how to calculate the biggest value that $R$ can get.

$\endgroup$

2 Answers 2

2
$\begingroup$

Le $w:=z-2i$, then for $0<|w|<4$, we have that $$f(z)=\frac{1}{w}\cdot \frac{1}{w+4i}=\frac{1}{4iw}\cdot \frac{1}{1-\frac{iw}{4}}=-\frac{i}{4w}\cdot \sum_{k=0}^{\infty}\left(\frac{iw}{4}\right)^k\\ =-\sum_{k=0}^{\infty}\left(\frac{i}{4}\right)^{k+1} w^{k-1} =-\sum_{k=-1}^{\infty}\left(\frac{i}{4}\right)^{k+2} (z-2i)^{k} $$

$\endgroup$
4
  • $\begingroup$ This looks like a really good solution, but how did you know so early that R = 4? I don't know how to spot this. $\endgroup$
    – armara
    Commented Oct 1, 2016 at 16:48
  • $\begingroup$ @armara Actually $4$ is the radius of convergence of the geometric series. Use the fact that $\sum_{k\geq 0} x^k=1/(1-x)$ iff $|x|<1$. In this case $x=iw/4$. $\endgroup$
    – Robert Z
    Commented Oct 1, 2016 at 16:50
  • $\begingroup$ I think I understand. So first I do the calculations, and when I see that I get $$\frac{1}{1-\frac{iw}{4}}$$, that's when I know that the radius of convergence is 4, am I right? $\endgroup$
    – armara
    Commented Oct 1, 2016 at 17:03
  • $\begingroup$ @armara Yes, that's right! $\endgroup$
    – Robert Z
    Commented Oct 1, 2016 at 17:34
1
$\begingroup$

We can use partial fraction expansion to write $f(z)$ as

$$f(z)=\frac{1}{i4}\left(\frac{1}{z-i2}-\frac{1}{z+i2}\right)$$

Note that the poles are separated by a distance $|i2-(-i2)|=4$. Therefore, let's write the Laurent series in the circle $|z-i2|<4$. The first term in that series is $\frac{1/i4}{z-i2}$. Let's expand the second term around $i2$.

$$\begin{align} \frac{1}{z+i2}&=\frac{1}{(z-i2)+i4}\\\\ &=\frac{1}{i4}\frac{1}{1+\frac{z-i2}{i4}}\\\\ &=\frac{1}{i4}\sum_{n=0}^\infty (-1)^n \left(\frac{z-i2}{i4}\right)^n\\\\ &=-\sum_{n=0}^\infty \left(\frac{i}{4}\right)^{n+1} (z-i2)^n \end{align}$$

Finally, we can write

$$f(z)=- \sum_{n=-1}^\infty \left(\frac{i}{4}\right)^{n+2} (z-i2)^n$$

$\endgroup$
2
  • $\begingroup$ I'm having problems when I use partial fraction with imaginary parts. I understand your solution, and I thank you for it, but I don't understand the first part where you conclude that $$f(z)=\frac{1}{i4}\left(\frac{1}{z-i2}-\frac{1}{z+i2}\right)$$. $\endgroup$
    – armara
    Commented Oct 1, 2016 at 17:14
  • $\begingroup$ Note that $$\frac{1}{i4}\left(\frac{1}{z-i2}-\frac{1}{z+i2}\right)=\frac1{i4}\left(\frac{(z+i2)-(z-i2))}{(z+i2)(z-i2)}\right)=\frac i4 \frac{i4}{z^2+4}=\frac{1}{z^2+4}=f(z)$$ $\endgroup$
    – Mark Viola
    Commented Oct 1, 2016 at 17:24

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .