Please check this complex integral

[ognik]

An old exam has: Evaluate $ \oint\frac{dz}{z(2z+1)} $, where the contour is a unit circle

This look good for the residue theorem, it has 2 simple poles at 0, $-\frac{1}{2}$
$ Res(f, 0)= \lim_{{z}\to{0}}z\frac{1}{z(2z+1)}=1$
$ Res(f, -\frac{1}{2})= \lim_{{z}\to{-\frac{1}{2}}}(z+\frac{1}{2})\frac{1}{z2(z+\frac{1}{2})}=-1$
$ \oint\frac{dz}{z(2z+1)} =2\pi i\sum Res(f) = 2 \pi [1 + (-1)] = 0$
Is that right?

 

[ThePerfectHacker]

Here is an easier suggestion. Use partial fractions first.

 

[Prove It]

An old exam has: Evaluate $ \oint\frac{dz}{z(2z+1)} $, where the contour is a unit circle

This look good for the residue theorem, it has 2 simple poles at 0, $-\frac{1}{2}$
$ Res(f, 0)= \lim_{{z}\to{0}}z\frac{1}{z(2z+1)}=1$
$ Res(f, -\frac{1}{2})= \lim_{{z}\to{-\frac{1}{2}}}(z+\frac{1}{2})\frac{1}{z2(z+\frac{1}{2})}=-1$
$ \oint\frac{dz}{z(2z+1)} =2\pi i\sum Res(f) = 2 \pi [1 + (-1)] = 0$
Is that right?

Looks good to me :)

 

[ognik]

Here is an easier suggestion. Use partial fractions first.

Hi TPH, I don't see how partial fractions are easier $ \implies \oint \left[ \frac{1}{z}-\frac{1}{z+\frac{1}{2}} \right] dz $ - I would still end up finding 2 residues? What am I missing?

 

[ThePerfectHacker]

Hi TPH, I don't see how partial fractions are easier $ \implies \oint \left[ \frac{1}{z}-\frac{1}{z+\frac{1}{2}} \right] dz $ - I would still end up finding 2 residues? What am I missing?

Use the following fact,
$$ \oint_{\Gamma} \frac{dz}{z-a} = \left\{ \begin{array}{ccc} 2\pi i & \text{ if } & a \text{ is inside the contour } \\
0 & \text{ if } & a \text{ is not inside } \end{array} \right. $$

You do not need to calculate anything. Only see if $a$ is inside or outside your contour.

 

[ognik]

Nice. So both are inside, therefore we get $2 \pi i - 2 \pi i$ = 0?