Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

how do I calculate $$\operatorname{Res}\left(\frac{1}{z^2 \cdot \sin(z))}, 0\right)$$ What is the order of the pole? $3$?

share|improve this question

2 Answers 2

Yes, the pole is of order 3. Thus

$$\begin{align}\mathrm{Res}_{z=0} \frac{1}{z^2 \sin{z}} &= \frac{1}{2!} \lim_{z \rightarrow 0} \frac{d^2}{d z^2} \left [ z^3 \frac{1}{z^2 \sin{z}} \right ] \\ &= \frac{1}{2}\lim_{z \rightarrow 0} \frac{d^2}{d z^2} \left ( \frac{z}{\sin{z}} \right )\\ &= \frac{1}{2} \lim_{z \rightarrow 0} [z (\csc^3{z} + \cot^2{z} \csc{z}) - 2 \cot{z} \csc{z}]\\ &= \lim_{z \rightarrow 0}\frac{1}{2} \left (\frac{2}{z^2}+\frac{1}{3}-\frac{2}{z^2} \right )\\ &= \frac{1}{6} \end{align}$$

share|improve this answer

You are right, $z=0$ is a pole of order 3 because $$ \lim_{z\to 0}\frac{z^3}{z^2\sin z}=\lim_{z\to 0}\frac{z}{\sin z}=1, $$ and since $$ \frac{1}{z^2\sin z}=\frac{1}{z^3(1-z^2/3!+o(z^3))}=\frac{1}{z^3}(1+\frac{z^2}{6}+o(z^3))=\frac{1}{z^3}+\frac{1}{6z}+o(1) $$ it follows that $$ \text{Res}(\frac{1}{z^2\sin z},0)=\frac16. $$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.