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**Cauchy Intergral Formula sin(i)??**

## Homework Statement

Circle of radius 2 centered at the origin oriented anticlockwise. Evaluate:

[tex]\int\frac{sin(z)}{z^{2} +1}[/tex]

## Homework Equations

I think I'm supposed to be using the Cauchy Integral Formula, so

[tex]\int\frac{f(z) dz}{z - z_{0}}[/tex] = 2[tex]\pi[/tex]

*i*f(z_{0})

## The Attempt at a Solution

I rewrote z[tex]^{2}[/tex] +1 = (z +

*i*)(z -

*i*) and took z[tex]_{0}[/tex] =

*i*, (suitable z[tex]_{0}[/tex] within the countour) so f(z) = [tex]\frac{sin(z)}{z + i}[/tex] .

Then 2[tex]\pi[/tex]

*i*f( [tex]_{0}[/tex] ) = 2[tex]\pi[/tex]

*i*[tex]\frac{sin(i)}{i + i}[/tex]

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([tex]\frac{\pi}{2}[/tex]) = 1 ? Is that correct or have I lost the plot somewhere?

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

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