# Thread: Complex Analysis - Residues at poles

1. ## Complex Analysis - Residues at poles

Let $f(z)=(1+z^2)/7z^3+50iz^2-7z$

(i). Determine the residue of $f$ at each of its poles
(ii). Deduce

$\int_0^{2 pi }(cost/25+7sint)dt=0$

2. Originally Posted by moolimanj
Let $f(z)=(1+z^2)/7z^3+50iz^2-7z$
What makes me angry is that you are in an advanced math class, and still did not learn how to use paranthesis

We have,
$f(z) = \frac{1+z^2}{7z^3+60iz^2-7z}$

The poles arise (possibly) when:
$7z^3+60iz^2 - 7z=0$

Now I am not going to do this problem because the solutions to that equation are messy, are you sure this is the problem?

3. Firstly, about the parenthesis - its my first time using the inbuilt equation editor (which I have never used before) so as long as it looked ok, i was happy rather than spending ages figuring it out. please accept my apologies. I'll have a play around with it to so that i can get it right next time.

Its definitely the problem. Like you say, its quite messy but rest assured it is whats required since this question gets a total of 8% of the paper.

4. I am sorry, it is not messy, I first read it as a 60 instead of a 50 which made it messy.

Let me continue the poles arise possibly when:
$7z^3+50iz^2-7z=0$
That is,
$z(7z^2+50iz-7)=0$
Thus,
$z=0 \mbox{ or }z=\pm 48i$.

It is in the special form $g(z)/h(z)$ where $g,h$ are holomorphic and $h'(c)\not = 0$.

So the residues can be computed as:
$\frac{1+z^2}{21z^2+100iz - 7}$ substitute for $z=0,\pm 48i$

5. Originally Posted by moolimanj
(ii). Deduce

$\int_0^{2 pi }(cost/25+7sint)dt=0$
$\int_0^{2\pi} \frac{\cos t}{25+7\sin t} dt = 0$

Consider the integral,
$\oint_{|z|=1} \frac{\frac{1}{2}\left( z + z^{-1} \right)}{25 + \frac{7}{2i}\left( z - z^{-1} \right)}\cdot \frac{1}{zi} \ dz$

Because if we integrate this using the parametrization $z=e^{it}$ on $0\leq t\leq 2\pi$ we get percisely the integral you have above.

Simpify the integral,
$-\frac{1}{4}\oint_{|z|=1}\frac{z^2+1}{7z^3+50iz-7z} \ dz$

Above we found that the poles where $z=0,\pm 48i$. Since we are working on the unit circle $|z|=1$ we see that $\pm 48i$ lie outside this circle. Thus, $z=0$ is the only pole.

So the integral value is,
$2\pi i \left( \frac{1}{-4} \right) \cdot \mbox{res}(f,0) = 0 + i \mbox{something}$

If we equal real and imaginaries we find that the real part is 0. Which is the value of the integral.