Math Help - Residue question

1. Residue question

Can someone tell me if my method is valid and my solution correct:

Calculate $\int_C \frac{sin z}{(z- \pi)^{10}} dz$

if the curve C is the circle:

a.
$|z- \pi| =1$
b.
$|z|=2$

Okay, the integrand is analytic on and inside the second curve, so the integral is zero there.

For the first curve I rewrite:

$\int_C \frac{sin z}{(z- \pi)^{10}} dz= \int_C \frac{e^{iz}}{2i(z- \pi)^{10}} dz - \int_C \frac{e^{-iz}}{2i(z- \pi)^{10}} dz$.

Then each term is easy to calculate using residues (noting the only singularity is at pi):

$\int_C \frac{e^{iz}}{2i (z- \pi)^{10}} dz - \int_C \frac{e^{-iz}}{2 i(z- \pi)^{10}} dz =$
$2 \pi i (\frac{d^{9}}{d^{9}z}\frac{e^{iz}}{9!2i}-\frac{d^{9}}{d^{9}z}\frac{e^{-iz}}{9!2i})$
$= 2 \pi (\frac{e^{iz}}{9!2} - \frac{-e^{-iz}}{9!2})$

Which evaluated at $z=\pi$ is $\frac{-2 \pi i}{9!}$

Does this work?

2. Originally Posted by igopogo
Can someone tell me if my method is valid and my solution correct:

Calculate $\int_C \frac{sin z}{(z- \pi)^{10}} dz$

if the curve C is the circle:

a.
$|z- \pi| =1$
b.
$|z|=2$

Okay, the integrand is analytic on and inside the second curve, so the integral is zero there.

For the first curve I rewrite:

$\int_C \frac{sin z}{(z- \pi)^{10}} dz= \int_C \frac{e^{iz}}{2i(z- \pi)^{10}} dz - \int_C \frac{e^{-iz}}{2i(z- \pi)^{10}} dz$.

Then each term is easy to calculate using residues (noting the only singularity is at pi):

$\int_C \frac{e^{iz}}{2i (z- \pi)^{10}} dz - \int_C \frac{e^{-iz}}{2 i(z- \pi)^{10}} dz =$
$2 \pi i (\frac{d^{9}}{d^{9}z}\frac{e^{iz}}{9!2i}-\frac{d^{9}}{d^{9}z}\frac{e^{-iz}}{9!2i})$
$= 2 \pi (\frac{e^{iz}}{9!2} - \frac{-e^{-iz}}{9!2})$

Which evaluated at $z=\pi$ is $\frac{-2 \pi i}{9!}$

Does this work?
Yes, but there may be an easier way to ascertain that last value. So, by Cauchy's integral theorem we have that $I=\int_{C}\frac{\sin(z)}{(z-\pi)^{10}}dz=2\pi i\left(\sin(z)\right)^{(9)}\mid_{z=\pi}$. But, notice that $\sin(z-\pi)=-\sin(z-\pi)$. But, remember that the Taylor series for $\sin(z-\pi)=\sum_{n=0}^{\infty}\frac{f^{(n)}(\pi)(z-\pi)^n}{n!}=\sum_{n=0}^{\infty}\frac{(-1)^n z^{2n+1}}{n!}$. Comparing this we see that $f^{(n)}(\pi)=\frac{(-1)^4}{(2(4)+1}=\frac{1}{9!}$. Thus, by previous comment we have that $I=-2\pi i\left(\sin(z)\right)^{(9)}\mid_{z=\pi}=-2\pi i\cdot\frac{1}{9!}$ which agrees with your answer.

3. Originally Posted by Drexel28
Yes, but there may be an easier way to ascertain that last value. So, by Cauchy's integral theorem we have that $I=\int_{C}\frac{\sin(z)}{(z-\pi)^{10}}dz=2\pi i\left(\sin(z)\right)^{(9)}\mid_{z=\pi}$. But, notice that $\sin(z-\pi)=-\sin(z-\pi)$. But, remember that the Taylor series for $\sin(z-\pi)=\sum_{n=0}^{\infty}\frac{f^{(n)}(\pi)(z-\pi)^n}{n!}=\sum_{n=0}^{\infty}\frac{(-1)^n z^{2n+1}}{n!}$. Comparing this we see that $f^{(n)}(\pi)=\frac{(-1)^4}{(2(4)+1}=\frac{1}{9!}$. Thus, by previous comment we have that $I=-2\pi i\left(\sin(z)\right)^{(9)}\mid_{z=\pi}=-2\pi i\cdot\frac{1}{9!}$ which agrees with your answer.
Okay, I owe you two thanks...one for validating my answer and another for the dope slap administered by showing me an easier way. Though it occurs to me that it should be even easier. After realizing that I can use Cauchy integral formula:

$f^{(9)}(\pi) = \frac{9!}{2 \pi i} \int_C \frac{sin(z)}{(z-\pi)^{10}} dz$

can't I simply evaluate $cos(\pi)$ as -1 (since cos is the 9th derivitive of sin), then use algebra to isolate the integral and get the answer?

4. Originally Posted by igopogo
Okay, I owe you two thanks...one for validating my answer and another for the dope slap administered by showing me an easier way. Though it occurs to me that it should be even easier. After realizing that I can use Cauchy integral formula:

$f^{(9)}(\pi) = \frac{9!}{2 \pi i} \int_C \frac{sin(z)}{(z-\pi)^{10}} dz$

can't I simply evaluate $cos(\pi)$ as -1 (since cos is the 9th derivitive of sin), then use algebra to isolate the integral and get the answer?
Of course