help integrating complex line integral How do i integrate $\int_c(z^2+1)^2dz$ along the arc of the cycloid $x=a(\theta-sin\theta),y=a(1-cos\theta)$ from the point where $\theta=0$ to $\theta=2\pi$.
I have tried putting the values by breaking z =x+iy, but it turned out to be too cumbersome. Can anyone offer a better approach? 
 A: I think your problem is more the line integral, i.e. integrating over the curve, rather than just integrating from one boundary to another, right?
You would solve this with help of the formula
$\int_C ds f(x) = \int_a^b dt f(x(t))\ |x'(t)|$
In your case $x(t)$ is the cycloid $x(\theta)= a((\theta-sin(\theta))+i(1-cos(\theta)))$
You also want to express your complex number $z$ in its polar representation as $z=r\ e^{i\theta}$. Remember $\mathbb{C} = \mathbb{R}\times\mathbb{R}$. Instead of the cartesian representation $\Re(z)\hat{e}_x + \Im(z) i\hat{e}_y$ you now choose the radius $r$ and the angle $\theta$ to represent your number in the complex plane.
This gives you the integral
$\int_0^{2\pi} d\theta \ (r²e^{2i\theta}+1)^2 x'(\theta)$
EDIT: Kevin is of course right, that the integral will be zero over the closed curve without singularities in the function to integrate. I just thought the problem is meant to be more general about line integrals.
A: Let $\gamma$ be the cycloid, since $\gamma(0)=\gamma(2π)=0$ and $(z^2+1)^2$ is analytic over the closed curve the line integral equals $0$.
