I want to make sure I remember how to do basic residue calculus problems for an exam, here's a problem:
Integrate $\int_C \frac{dz}{1+z^2}$ if
(a) C is the circumference given by $\mid z - i\mid = 1$ and if
(b) C is the circumference given by $\mid z \mid =2$
Answer:
(a) C is a circle of radius 1 centered at $i$, and $f(z)=\frac{1}{(z-i)(z+i)}$ so it has poles in $\pm i$. Only the positive one is included in the contour. We have $Res(f,i)=\lim_{z\rightarrow i}(z-i)\frac{1}{(z-i)(z+i)}=\frac{1}{2i}$
So $\int_C \frac{dz}{1+z^2}=2\pi i \frac{1}{2i}=\pi$
(b) C is a cirle of radius 2 centered at the origin, it includes both the positive and negative poles of $f(z)$.
To find the residue for the negative pole: $\lim_{z\rightarrow -i}(z+i)\frac{1}{(z+i)(z-i)}=-\frac{1}{2i}$
So $\int_C \frac{dz}{1+z^2}=2\pi i (\frac{1}{2i}-\frac{1}{2i})=0$
Does everything check out? Much appreciated.
People seem to be happy with this, I'm posting it as an answer instead of shutting down the post.
(a) C is a circle of radius 1 centered at $i$, and $f(z)=\frac{1}{(z-i)(z+i)}$ so it has poles in $\pm i$. Only the positive one is included in the contour. We have $Res(f,i)=\lim_{z\rightarrow i}(z-i)\frac{1}{(z-i)(z+i)}=\frac{1}{2i}$
So $\int_C \frac{dz}{1+z^2}=2\pi i \frac{1}{2i}=\pi$
(b) C is a cirle of radius 2 centered at the origin, it includes both the positive and negative poles of $f(z)$.
To find the residue for the negative pole: $\lim_{z\rightarrow -i}(z+i)\frac{1}{(z+i)(z-i)}=-\frac{1}{2i}$
So $\int_C \frac{dz}{1+z^2}=2\pi i (\frac{1}{2i}-\frac{1}{2i})=0$