Evaluate $\int_{|z|=3}\frac{dz}{z^3(z^{10}-2)}$

Question

$$\int_{|z|=3}\frac{dz}{z^3(z^{10}-2)}$$

There are singularities at $z=0$ and $z^{10}=2\iff z=\sqrt[10]{2}e^{\frac{i\pi k}{5}} \text{ where } k=0,1,...9$ so we need to find $10$ residues? or have I got it wrong?

score 4 · Answer 1 · answered Jul 15 '18 at 17:13

4

Since there are no singularities outside $|z|=2^{1/10}$, the Cauchy Integral Theorem says that $$ \int_{|z|=3}\frac{\mathrm{d}z}{z^3(z^{10}-2)} =\int_{|z|=r}\frac{\mathrm{d}z}{z^3(z^{10}-2)} $$ for all $r\gt2^{1/10}$. As $r\to\infty$, the integral on the right obviously vanishes.

answered Jul 15 '18 at 17:13

robjohn

345,667

Why can’t we do the same with $f(z)=\frac{dz}{z^2+1}$ at $|z|=2$ – gbox Jul 15 '18 at 17:27
@gbox: We can. Why do you think we can't? – robjohn Jul 15 '18 at 18:18
So $\int_{|z|=2}\frac{dx}{z^2+1}=0$ and not $2\pi i (res(f,i)+res(f,-i))$? – gbox Jul 15 '18 at 18:34
the residue of $\frac1{z^2+1}$ at $i$ is $-\frac i2$. The residue at $-i$ is $\frac i2$, so the sum of the residues is $0$. – robjohn Jul 15 '18 at 19:17
1

@gbox If you calculate those residues and that doesn't make you wish you'd calculated them before making that comment then you calculated them wrong. – David C. Ullrich Jul 15 '18 at 19:17

David C. Ullrich · Answer 2 · 2018-07-15T17:30:09.977

This is easy by symmetry, without calculating any residues.

Say $f(z)=\frac1{z^3(z^{10}-2)}$ and let $\omega=e^{2\pi i/10}$. Note that $$f(\omega z)=\omega^{-3}f(z).$$It follows that the integral is $0$.

Hint regarding the relevant calculations: Say the integral is $I$. Writing slightly informally, $$I=\int_{|z|=3}f(\omega z)\,d(\omega z) =\int_{|z|=3}\omega^{-3}f(z)\omega\,dz=\omega^{-2}I;$$since $\omega^{-2}\ne1$ this shows $I=0$.

(The "informal" part is the first equality above. To justify it, note that if $t\mapsto\gamma(t)$ is a parametrization of $|z|=3$ then $t\mapsto\omega\gamma(t)$ is also a parametrization.)

Empty · Answer 3 · 2018-07-25T15:35:38.500

2

Hint : Sum of the residues at all singularities including infinity is equal to zero. So it is sufficient to evaluate the singularity of $f$ at $z=\infty$, where $\displaystyle f(z)=\frac{1}{z^3(z^{10}-2)}$ and then apply Cauchy's Residue Theorem.

First line says that, If $f$ has poles at $z=a_i(i=1,2,\cdots ,n)$ then $\displaystyle \sum_{i=1}^n \text{Res}(f;a_i) +\text{Res}(f;\infty)=0$.

edited Jul 25 '18 at 15:35

answered Jul 15 '18 at 16:24

Empty

13,012

Why "it is sufficient to evaluate the singularity of $f$ at $z=\infty$" is there a theorem about it? – gbox Jul 15 '18 at 16:28
1

Read the previous line again – Empty Jul 15 '18 at 16:29
@gbox See the edited answer. – Empty Jul 15 '18 at 16:35

score 0 · Answer 4 · answered Jul 15 '18 at 16:33

Actually, that would mean $11$ residues, but since $\operatorname{Res}\left(\frac1{z^3(z^{10}-2)},0\right)=0$…

On the other hand, if $z_k=\sqrt[10]2e^{\frac{k\pi i}5}$ ($k\in\{0,1,\ldots,9\}$), then\begin{align}\operatorname{Res}\left(\frac1{z^3(z^{10}-2)},z_k\right)&=\frac1{{z_k}^310{z_k}^9}\\&=\frac1{10{z_k}^{12}}\\&=\frac{\sqrt[10]2^{-12}}{10}\exp\left(-\frac{12k\pi i}5\right)\\&=\frac{\sqrt[10]2^{-12}}{10}\exp\left(-\frac{12\pi i}5\right)^k.\end{align}So, what you have to sum is a gemetric progression.

Evaluate $\int_{|z|=3}\frac{dz}{z^3(z^{10}-2)}$

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