1. ## Complex Integration

I'm stuck on a question of complex integration; there doesn't seem to be any way to integrate the function, so I imagine there must be a 'trick' somewhere, but I can't work it out Here's the problem:

Compute
$\oint \frac{log(z)}{z^2 + 9}dz$
around $|z-4i|=3$.

I parametrized the path as
$\gamma (t) = 4i + 3e^{it},\ 0 \leq t \leq 2\pi$
but then I end up trying to integrate
$\int_0^{2\pi} \frac{log(4i+3e^{it})3ie^{it}}{(4i+3e^{it})^2 + 9}dt$
which I can't work out. Thanks for any help

2. Why not use the residue theorem?

3. Originally Posted by Defunkt
Why not use the residue theorem?
I thought there have to be poles inside the contour. $f(z)$ has poles at $z=\pm 3$, which aren't inside $\gamma (t)$.

4. It actually has poles at $z= \pm 3i$

:P

EDIT: And if it had no poles then it would be analytic inside that domain, which means that the integral would've been 0.

5. Originally Posted by Defunkt
It actually has poles at $z= \pm 3i$

:P

EDIT: And if it had no poles then it would be analytic inside that domain, which means that the integral would've been 0.
Damn.. Such a stupid mistake Thanks for pointing it out!

6. Alright, so how do I go about calculating the residue? Don't I still need to integrate $f(x)$ to work out $Res(f, z_1)$ where $z_1 = 3i$?

7. No.
Since $z=3i$ is a simple pole, you know that

$Res(f, 3i) = \lim_{z \to 3i}(z-3i)f(z)$

8. Originally Posted by Defunkt
No.
Since $z=3i$ is a simple pole, you know that

$Res(f, 3i) = \lim_{z \to 3i}(z-3i)f(z)$
How do we know that $z = 3i$ is a simple pole?

9. How do we know that z = 3i is a simple pole?
Factor the denominator of the original integrand. The factor z - 3i appears to the first power. That implies, by definition, pretty much, that it's a simple pole.

10. Originally Posted by Ackbeet
Factor the denominator of the original integrand. The factor z - 3i appears to the first power. That implies, by definition, pretty much, that it's a simple pole.
Man, I wish the definition I have was that simple... Why can't my lecturer's explain it like that?! Thanks everyone

11. Well, you gotta make sure the numerator doesn't also go to zero or do something else equally strange at the same location. The precision necessary to correctly define a simple pole means you're probably going to be doing something similar to what your lecturers have. However, this is an intuitive way to think of poles, and it works in quite a few cases.

12. In a slightly related question, compute
$\oint \frac{\sin (2z)}{z^{15}}$
around $|z| = 2$.

There's obviously a pole at $z=0$ but is that a simple pole, as the numerator will be zero at that point...

13. The numerator will be zero at that point, correct. However, the denominator has a 15th order zero at the origin. You can show that the sin function has a first-order zero at the origin. Therefore, the integrand has a 14th order pole at the origin. Make sense?

14. Originally Posted by Ackbeet
The numerator will be zero at that point, correct. However, the denominator has a 15th order zero at the origin. You can show that the sin function has a first-order zero at the origin. Therefore, the integrand has a 14th order pole at the origin. Make sense?
Cool, that makes sense. So I can work out the residue at $z=0$, by using
$b_{1} = \frac{1}{2\pi i} \int_\gamma f(\zeta )d\zeta$
right? I did that and after a bit of manipulation, arrived at
$\frac{1}{2^{15} \pi } \int_{0}^{2\pi }\frac{\sin{4e^{it}}}{e^{14it}}$
which I can't integrate. I've got no idea how to do it..

15. I would just evaluate the residue using this formula.

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