Image of a closed curve under $w=z^2$. I have the curve:
$$r=2(1+ \cos \theta), \ \theta \in [0,2\pi)$$
in polar coordinates on the complex $z$ plane, and I want to find the image of this curve under the square function $w=f(z)=z^2$. First of all, I have no idea how to attempt to recognize this curve without resorting to a computer generated graph, it seems to be some fourth degree curve. But perhaps it's not important what it looks like, maybe it has a simple image, so what I tried was to say that every point on the curve is of the form $$z(\theta)=2(1+\cos \theta ) e^{i \theta}, \ \theta \in [0,2\pi),$$
so squaring turns each point into 
$$w(\theta)=4(1+\cos\theta)^2 e^{2i \theta}, \ \theta \in [0,2\pi)$$
But I have no idea what I'm supposed to do now. Is there a way to simplify this? Is there a better approach? I can't even write an equation of the form $r=g(\theta)$, because its a $e^{2i \theta}$ not $e^{i \theta}$...
 A: Resulting curve will loop twice around the origin therefore you cannot express your curve as function $r=g(\theta)$. You got the answer right that 
$$
w(\theta)=4(1+\cos\theta)^2 e^{2i \theta}, \ \theta \in [0,2\pi)
$$
You should look at this as parametric equation, so as $w : [0,2\pi)\rightarrow \mathbb{R}^2$
$$
w(\theta)=4(1+\cos\theta)^2 (\cos(2\theta),\sin(2\theta))
$$
To make a picture of this curve just plug $0,\frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4},\dots$ into $w(\theta)$ and than connect those points in some sencible manner. 
Here is the plot done with W|A:

You can cut solution into two pieces $r = g_1(\theta), r=g_2(\theta)$ where
$$
g_1(\theta) = \left(2 + 2 \cos{\frac{\theta}{2}}\right)^2
$$
with graph

$$
g_2(\theta) = \left(2 + 2 \cos{\frac{\theta}{2} + \pi} \right)^2
$$
with graph

A: You should recognize that $r=|z|$. In this case, using $2\cos{\theta} = e^{i \theta}+e^{-i \theta}$, you get that $z=\left ( 1+e^{i \theta}\right)^2$.  The image of this under $w=z^2$ is $w=\left ( 1+e^{i \theta}\right)^4$, and the polar curve that results is, as expected, $r = (2+2 \cos{\theta})^2$.
