[SOLVED] Complex integral

**arbolis** · October 19th 2009, 04:11 PM

Let $C$ be the unit circle with the positive sense.
Evaluate $\int _C \frac{dz}{\sin z}$ .
My attempt :

Ok so first I notice that $\sin (z)=0$ only when $z=0$ (at least in the circle, if I'm not wrong).
So there is one singularity in the domain of integration.
My idea was to determine the order of the pole (I'm all confused with poles and singularities) of $\frac{1}{\sin z}$ at $z=0$ .
I found the Laurent series of $\frac{1}{f(z)}$ to be worth $1-\frac{z^3}{3!}+\frac{z^5}{5!}-...=1-z^3 \left ( \frac{1}{3!}-\frac{z^2}{5!}+... \right )$ thus the zero of $\frac{1}{f(z)}$ is of order 3, hence the order of the pole of $f$ is of order $3$ .
This implies that the integral is worth $3\cdot 2 \pi i= 6\pi i$ .
Am I right?
(sorry for all these questions, tomorrow I'm going to ask help at university since I've a lecture. I might be asking here at night though

)

**chisigma** · October 19th 2009, 08:04 PM

The $\sin (*)$ function has the following expansion as 'infinite product'...

$\sin z = z \cdot \prod_{n=1}^{\infty} (1-\frac{z^{2}}{\pi^{2} n^{2}})$ (1)

... so that is...

$\frac{1}{\sin z} = \frac{1}{z}\cdot \prod_{n=1}^{\infty} \frac{1}{1-\frac{z^{2}}{\pi^{2} n^{2}}}$ (2)

From (2) it is evident that $\frac{1}{\sin z}$ in $z=0$ has a simple pole and its residue is...

$r_{0}= \lim_{z \rightarrow 0} \frac{z}{\sin z}=1$ (3)

But $z=0$ is yhe only pole of $f(*)$ inside the unit circle so that is...

$\int_{C} \frac{dz}{\sin z} = 2\pi i$ (4)

Kind regards

$\chi$ $\sigma$

**mr fantastic** · October 19th 2009, 09:22 PM

Originally Posted by arbolis

Let $C$ be the unit circle with the positive sense.
Evaluate $\int _C \frac{dz}{\sin z}$ .
My attempt :

Ok so first I notice that $\sin (z)=0$ only when $z=0$ (at least in the circle, if I'm not wrong).
So there is one singularity in the domain of integration.
My idea was to determine the order of the pole (I'm all confused with poles and singularities) of $\frac{1}{\sin z}$ at $z=0$ .
I found the Laurent series of $\frac{1}{f(z)}$ to be worth $1-\frac{z^3}{3!}+\frac{z^5}{5!}-...=1-z^3 \left ( \frac{1}{3!}-\frac{z^2}{5!}+... \right )$ thus the zero of $\frac{1}{f(z)}$ is of order 3, hence the order of the pole of $f$ is of order $3$ .
This implies that the integral is worth $3\cdot 2 \pi i= 6\pi i$ .
Am I right?
(sorry for all these questions, tomorrow I'm going to ask help at university since I've a lecture. I might be asking here at night though

)

Note that $\lim_{z \rightarrow 0} z f(z) = \lim_{z \rightarrow 0} \frac{z}{\sin z} = 1$ and so $z = 0$ is a simple pole (see also the thread from a couple of day ago where this sort of thing was discussed). Knowing that the pole is simple, this calculation also tells you that $\text{Res}( f(z), z = 0) = 1$ .

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