How do I expand $|z_{1} - z_{2}|^2$?

Question

I'm trying to expand $|z_{1} - z_{2}|^2$ where $z_{1}$ and $z_{2}$ are complex numbers.

Is it exactly the same as with real numbers $z_{1}$ and $z_{2}$?

Answer 1

Also take a look at user57404 and Will Jagy's comments. To wrap this up in an answer, you have

$$z\bar{z}=|z|^2$$

Thus,

$$|z_1-z_2|^2=(z_1-z_2)\overline{z_1-z_2}=(z_1-z_2)(\bar{z_1}-\bar{z_2})$$

as complex conjugation distributes over sums and products. Now, if you want to go further, you may multiply this out:

$$(z_1-z_2)(\bar{z_1}-\bar{z_2})=z_1\bar{z_1}-z_1\bar{z_2}-z_2\bar{z_1}+z_2\bar{z_2}=|z_1|^2+|z_2|^2-(z_1\bar{z_2}+z_2\bar{z_1})=|z_1|^2+|z_2|^2-2\mathrm{Re}(z_1\bar{z_2})$$

So structurally, there is some similarity to the real case.

Answer 2

Using polar coordinates, write $z_1=\rho e^{i\phi}$ and $z_2=re^{i\theta}$ where $\rho=|z_1|$, $r=|z_2|$, $\phi=\arg(z_1)$, and $\theta=\arg(z_2)$. Then we see that

$$\begin{align} |\rho e^{i\phi}-re^{i\theta}|^2&=|\left(\rho \cos(\phi)-r\cos(\theta)\right)+i\left(\rho\sin(\phi)-r\sin(\theta)\right)|^2\\\\ &=\left(\rho \cos(\phi)-r\cos(\theta)\right)^2+\left(\rho\sin(\phi)-r\sin(\theta)\right)^2\\\\ &=\rho^2+r^2-2\rho r\cos(\theta-\phi)\\\\ &=|z_1|^2+|z_2|^2-2|z_1||z_2|\cos(\arg(z_1)-\arg(z_2)) \end{align}$$