Equation of cardioid

Question

Can someone please explain to how

$$c = \frac{1}{2} e^{i \theta}- \frac{1}{4} e^{2 i \theta}, \quad 0 \leq \theta \leq 2 \pi$$

does represent an equation of cardioid $r = \frac{1}{2} - \frac{1}{4} \cos(\theta)$, $0\leq \theta \leq 2 \pi$?

I tried to write $\cos(\theta) = \frac{e^{i \theta} + e^{-i \theta}}{2}$, but couldn't prove it is equal to $\frac{1}{2} e^{i \theta}- \frac{1}{4} e^{2 i \theta}$. Thanks in advance.

Answer 1

Plot the two functions. You will then note that:

1. The formula for the radius is wrong. In a cardioid the radius is zero at $\theta=0$. Try $$r=\frac12(1-\cos\theta)$$
2. The complex representation is a carioid shifted along the real axis by $1/4$

$$x_c=\frac12\cos\theta-\frac14\cos2\theta=\frac12\cos\theta-\frac12\cos^2\theta+\frac14=\frac12(1-\cos\theta)\cos\theta+\frac14\\y_c=\frac12\sin\theta-\frac14\sin2\theta=\frac12\sin\theta-\frac12\sin\theta\cos\theta=\frac12(1-\cos\theta)\sin\theta $$

Answer 2

For a given value of real $c1$ ( c,s are cos t ,sin t)

the 2D $t$ parametrization is

$$ X(t)= c1 (c/2-(c^2-s^2)/4, (1-c)s/2) $$

For complex $c1= e^ \alpha, $ there is a rotation of the full cardoid by angle $\alpha$.

Answer 3

It doesn't.

This is $\frac{1}{2}e^{i \theta}-\frac{1}{4}e^{2i\theta}$ with $0\le \theta \le 2\pi$ :

and this is $r=\frac{1}{2}-\frac{1}{4}\cos \theta$ with $0 \le \theta \le 2\pi$ :