- #1
Meggle
- 16
- 0
Cauchy Intergral Formula sin(i)??
Circle of radius 2 centered at the origin oriented anticlockwise. Evaluate:
[tex]\int\frac{sin(z)}{z^{2} +1}[/tex]
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]if(z_{0})
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]if( [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.)
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]if(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]if( [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.)
Last edited: