Contour Integration: Calculating Residues and Solving Integrals

[quasar987]

[SOLVED] contour integration

Homework Statement

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

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

The Attempt at a Solution

Well,

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

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

Now I need to parametrize C_rho. Take

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

We have

$$\frac{d\gamma}{dt}=i\sqrt{2}\pi e^{i2\pi t}$$

So that

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

Now what??

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

 

[Despondent]

Why don't you just calculate the residue directly?

 

[quasar987]

Refresh my memory please?

 

[Despondent]

The singularity at $$e^{\frac{{i\pi }}{4}}$$ 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

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

In the calculation you can simplify the algebra a little by using $$z_0 ^4 = - 1$$.

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

 

[quasar987]

Nice, thank you so much.