0
$\begingroup$

How can i find the principal part of $ (2z+3)/(z+2)^2$ ? I cannot seem to be able to expand this series. I want to know the residue at the pole of this function as well.

$\endgroup$
3
  • $\begingroup$ Unless I'm missing something, the answer is right in front of you: the principal (= negative power part of the Laurent series) at $z=-2$ is $\color{blue}{0}\cdot (z+2)^{-1}+3\cdot (z+2)^{-2}$ so the residue of $f$ at $-2$ is $\color{blue}{0}$... $\endgroup$
    – StackTD
    Commented Mar 23, 2017 at 10:15
  • $\begingroup$ I just edited the question, could you take a look once more? $\endgroup$
    – Nipesh Kc
    Commented Mar 23, 2017 at 10:18
  • $\begingroup$ Alright, see answer below. $\endgroup$
    – StackTD
    Commented Mar 23, 2017 at 10:23

3 Answers 3

2
$\begingroup$

Rewrite as follows to get the principal part of the Laurent series at $z=-2$ in standard form: $$\frac{2z+\color{red}{3}}{\left(z+2\right)^2} =\frac{2z+\color{red}{4-1}}{\left(z+2\right)^2} = \frac{2\left(z+2\right)}{\left(z+2\right)^2}-\frac{1}{\left(z+2\right)^2} = \frac{\color{blue}{2}}{z+2}-\frac{1}{\left(z+2\right)^2}$$ So the residue (coefficient of $\left(z+2\right)^{-1}$) is $\color{blue}{2}$.

$\endgroup$
2
$\begingroup$

Hint:

Use the substitution $u=(z+2)$ so that the function becomes: $$ \frac{2u-1}{u^2}=\frac{2}{z+2}-\frac{1}{(z+2)^2} $$.

$\endgroup$
2
$\begingroup$

If you only want to know the residue at the pole$-2$, then use the following rule:

If $f$ has a pole of order $2$ at $a$ and if $g(z):=(z-a)^2f(z)$, then

$$ res(f;a)= \lim_{z \to a}g'(z).$$

In your case we have $g(z)=2z+3$, hence $g'(z)=2$ and therefore

$$ res(f;-2)= 2.$$

$\endgroup$

You must log in to answer this question.

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