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

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 • But in Ron Gordon's answer he expressed it as $r=g(\theta)$. Or did he just "cut out" the bit that overlaps? Sep 30, 2013 at 17:36
• well because his answer is wrong $(2+2\cos \theta)^2 e^{i\theta} \neq \left( 1+ e^{i \theta} \right)^4$.
– tom
Sep 30, 2013 at 18:01
• I edited my answer, you can cut the solution into two pieces.
– tom
Sep 30, 2013 at 18:16

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, $r = (2+2 \cos{\theta})^2$.

• I don't see how to get $r = (2+2 \cos{\theta})^2$ from $w=\left ( 1+e^{i \theta}\right)^4$. Sep 30, 2013 at 17:32
• @DepeHb: $$(1+\cos{\theta})^2+\sin^2{\theta} = 2 + 2 \cos{\theta}$$ Also, $|z|^4 = (|z|^2)^2$. Sep 30, 2013 at 17:38
• Are you aware that $(2+2\cos \theta)^2 e^{i\theta} \neq \left( 1+ e^{i \theta} \right)^4$? So your answer does not make any sense to me.
– tom
Sep 30, 2013 at 18:03
• @tom: Yeah, but the magnitudes are equal. Sep 30, 2013 at 18:26
• @tom: the equality holds for the reason I stated above. Please explain why you disagree. Sep 30, 2013 at 19:03