Contour Integration: Calculating Residues and Solving Integrals

[quasar987]

[SOLVED] contour integration

Homework Statement

I'm really rusty with this. I need to calculate

[tex]2\pi i Res\left(\frac{1}{1+z^4},e^{i\pi/4}\right)[/tex]

The Attempt at a Solution

Well,

[tex]2\pi iRes\left(\frac{1}{1+z^4},e^{i\pi/4}\right)=\int_{C_{\rho}}\frac{1}{1+z^4}dz[/tex]

where [tex]C_{\rho}[/tex] is a little circle of radius rho centered on [tex]e^{i\pi/4}[/tex], on which there are no singularities. Fine, so let's take [tex]\rho=1/\sqrt{2}[/tex].

Now I need to parametrize C_rho. Take

[tex]\gamma(t)=e^{i\pi/4}+\frac{e^{i2\pi t}}{\sqrt{2}} \ , \ \ \ \ 0\leq t < 1[/tex]

We have

[tex]\frac{d\gamma}{dt}=i\sqrt{2}\pi e^{i2\pi t}[/tex]

So that

[tex]\int_{C_{\rho}}\frac{1}{1+z^4}dz=\int_0^1\frac{i\sqrt{2}\pi e^{i2\pi t}}{1+(e^{i\pi/4}+\frac{1}{\sqrt{2}}e^{i2\pi t})^4}dt[/tex]

Now what??

I believe the answer is supposed to be [tex]-e^{i\pi/4}/4[/tex]

 

[Despondent]

Why don't you just calculate the residue directly?

 

[quasar987]

Refresh my memory please?

 

[Despondent]

The singularity at [tex]e^{\frac{{i\pi }}{4}} [/tex] is a simple pole and the numerator of the integrand is non zero and holomorphic at the singularity. So you can calculate the residue by using the formula

[tex]
{\mathop{\rm Re}\nolimits} s\left( {\frac{1}{{z^{^4 } + 1}}} \right) = \frac{{1_{z = z_0 } }}{{\frac{d}{{dx}}\left( {z^4 + 1} \right)}}_{z = z_0 } ,z_0 = e^{\frac{{i\pi }}{4}}
[/tex]

In the calculation you can simplify the algebra a little by using [tex]z_0 ^4 = - 1[/tex].

The result should coincide with the answer that you expected to get as indicated in your original post.

 

[quasar987]

Nice, thank you so much.