Real part of a quotient Is there some fast way to know the real part of a quotient?
$$\Re\left(\frac{z_1}{z_2}\right)$$
$z_i\in \mathbb{C}$
 A: Given $z_1=a+bi$ and $z_2=c+di$ then:$$\Re\left(\frac{z_1}{z_2}\right)=\frac{ac+bd}{c^2+d^2}$$
A: If you don't have $z_1$ and $z_2$ in a "nice" form eg some of the values of $e^{i\theta}$ for various $\theta$, you could use $1/2(z+\bar{z})=\Re{(z)}$.
So you'd get:
$\Re(\frac{z_1}{z_2})=\frac{z_1\bar{z_2}+\bar{z_1}z_2}{2z_2\bar{z_2}}$
for
$z_1=r_1e^{i\theta_1}$
$z_2=r_2e^{i\theta_2}$
$\Re(\frac{z_1}{z_2})=\frac{r_1}{r_2}cos(\theta_1-\theta_2)$
Just as an alternative, although I suspect you wanted something like user130512's post.
A: At user130512's request in a comment, Euler's form. (I really suspect the calculation is going to get bad).
Let $z_1 = r_1e^{i\theta_1}$ and $z_2 = r_2e^{i\theta_2}$.
Now $\frac{z_1}{z_2} = \frac{r_1}{r_2}e^{i(\theta_1 - \theta_2)}$
Assuming $z_1 = a + ib, z_2 = c+id$
$$\Re\left(\frac{z_1}{z_2}\right) = \frac{r_1}{r_2}\cos\left(\arctan\left(\frac{b}{a}\right) - \arctan\left(\frac{d}{c}\right)\right)$$
Disclaimer This is for pure amusement. No practical point in doing this. 
