Cauchy Intergral Formula sin(i)?

[Meggle]

Cauchy Intergral Formula sin(i)??

Homework Statement

Circle of radius 2 centered at the origin oriented anticlockwise. Evaluate:
\int\frac{sin(z)}{z^{2} +1}

Homework Equations

I think I'm supposed to be using the Cauchy Integral Formula, so
\int\frac{f(z) dz}{z - z_{0}} = 2\piif(z_{0})

The Attempt at a Solution

I rewrote z^{2} +1 = (z + i)(z - i) and took z_{0} = i , (suitable z_{0} within the countour) so f(z) = \frac{sin(z)}{z + i} .

Then 2\piif( _{0} ) = 2\pii \frac{sin(i)}{i + i}

But what do I do with sin(i)? Can I take i in polar form on my real/imaginary axis and say sin(i) = sin(\frac{\pi}{2}) = 1 ? Is that correct or have I lost the plot somewhere?

(Sorry, I never seem to get the Latex quite right.)

 

[snipez90]

Yes it should be fine to leave it as sin(i). But clearly sin(i) =/= 1, so I'm not sure what you were trying to do at the end. Using sin(z) = (e^iz - e^(-iz)) / 2i should clarify this. The rest of your work seems fine.

 

[gomunkul51]

Also, you may use the Residue Theorem to solve this problem.

1. Both i and -i are inside the circle of radius 2 (centered at origin), so you have 2 different singularities inside your closed contour.

2. sin(z) = (exp(iz) - exp(-iz))/2i

 

[Meggle]

Also, you may use the Residue Theorem to solve this problem.

1. Both i and -i are inside the circle of radius 2 (centered at origin), so you have 2 different singularities inside your closed contour.

2. sin(z) = (exp(iz) - exp(-iz))/2i

O yes, I can't use Cauchy because there's more than one singular point. Cheers.

 

[snipez90]

Ah yes, stupid oversight, since f was clearly not analytic on {z | |z| < 2}. So you have to use residue calculus, which is fine.

*EDIT* Actually, you can still use Cauchy (especially if you don't know the residue theorem). But you have to apply http://en.wikipedia.org/wiki/Cauchy%27s_integral_theorem" first before applying Cauchy's integral formula (this is slightly confusing but everything else is based on Cauchy's theorem). This is the correct way to deal with multiple singularities via the Cauchy integral formula, though on the few complex integrals I have looked at, it happened that I ditched this method in favor of residue calculus because the former is tedious.