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 $e^{i \theta}$...

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