I noticed that the complex function $$f(z) = \frac{2}{(z+i)^2}$$ seems to map the real line onto the cardioid given by the polar equation: $$r = 1- \cos(\theta).$$ I was wondering if there is a simple explanation for this? I did note that, for $t \in \mathbb{R}$, one can write $$f(t) = \frac{2}{(t+i)(t-i)} \cdot \frac{t-i}{t+i} = r(t) u(t)$$ where $r(t) = \frac{2}{t^2+1}$ maps $\mathbb{R} \to [0,2]$ and $u(t) = \frac{t-i}{t+i}$ is Cayley transform which maps $\mathbb{R}$ to the unit circle.
Is there an elegant way to continue from here?
Writing $t = \cot\varphi$, we obtain
$$\begin{align} \frac{2}{t^2+1}\frac{t-i}{t+i} &= \frac{2}{\cot^2\varphi + 1}\frac{\cot\varphi - i}{\cot\varphi+i}\\ &= \frac{2\sin^2\varphi}{\cos^2\varphi+\sin^2\varphi}\frac{\cos\varphi - i\sin\varphi}{\cos\varphi + i\sin\varphi}\\ &= 2\sin^2\varphi \frac{e^{-i\varphi}}{e^{i\varphi}}\\ &= (1 - \cos (2\varphi))e^{-2i\varphi}. \end{align}$$
Rename $\theta = -2\varphi$.
The simple explanation is that the inversion $z \mapsto \frac{1}{z+i}$ maps the real line to the circle $\lvert w + \frac{i}{2}\rvert = \frac{1}{2}$, and squaring maps circles passing through the origin to cardioids.