Prove $|z^2| = |z|^2$ using Polar Coordinates

Question

I'm having trouble proving the above theorem.

Could someone explain to me how I could prove it using Polar Coordinates?

I've done so without them, but I just can't figure out how you're meant to ...

Answer 1

Recall that $|e^{i\theta}| = 1$ for all $\theta\in \mathbb{R}$. So if $z = re^{i\theta}$, we have $|z| = |r||e^{i\theta}| = |r| \Rightarrow |z|^2 =|r|^2 $. Likewise you can check that $|z^2| = |r^2||e^{2i\theta}| = |r^2| = |r|^2$.

Answer 2

Let $z = r(\cos(\theta) + i\sin(\theta))$, where $r\in[0,+\infty)$ and $\theta\in[0,2\pi)$.

Then one gets the desired result as follows: \begin{align*} |z^{2}| & = |(r(\cos(\theta) + i\sin(\theta))^{2}|\\\\ & = |r^{2}(\cos(\theta) + i\sin(\theta))^{2}|\\\\ & = |r^{2}(\cos^{2}(\theta) - \sin^{2}(\theta)) + 2i\sin(\theta)\cos(\theta))|\\\\ & = |r^{2}(\cos(2\theta) + i\sin(2\theta))|\\\\ & = \sqrt{(r^{2}\cos(2\theta))^{2} + (r^{2}\sin(2\theta))^{2}}\\\\ & = \sqrt{r^{4}(\cos^{2}(2\theta) + \sin^{2}(2\theta))}\\\\ & = r^{2} = |z|^{2} \end{align*}

Hopefully this helps!

BONUS

Given two complex numbers $z\in\mathbb{C}$ and $w\in\mathbb{C}$, the following property holds: \begin{align*} |zw| = |z||w| \end{align*}

Proof

Indeed, one has: \begin{align*} |zw|^{2} & = (zw)(\overline{zw})\\\\ & = zw\overline{z}\overline{w}\\\\ & = (z\overline{z})(w\overline{w})\\\\ & = |z|^{2}|w|^{2} \end{align*} whence we conclude $|zw| = |z||w|$. Now it suffices to take $z = w$, and we are done.

Answer 3

The beauty of polar coordinates is that $\sin^2 \theta + \cos^2 \theta$ is always equal to one. And for any two numbers $s,c$ so that $s^2 + c^2 = 1$ there will always be an angle $\phi$ so that $s =\sin \phi$ and $t=\cos \phi$ (simply let $\phi = \arctan \frac sc$)

So if we have a non-zero complex number $z= a + bi$ we will have $|z| = \sqrt{a^2 + b^2}> 0$ and we can define $r = |z|$ and think of that as the magnitude (size) of $z$. Now if we factor out $r$ from $a+bi$, we get $z = a+bi = r(\frac ar + \frac bri)$.

Now here's the beauty of it all! $(\frac ar)^2 + (\frac br)^2 = \frac {a^2}{\sqrt{a^2 + b^2}^2} + \frac {b^2}{\sqrt{a^2 + b^2}^2}=\frac {a^2 + b^2}{a^2 +b^2} = 1$. Always! and so there is an angle $\theta = \arctan \frac ba$ so that we can write $z = a+bi = r(\cos \theta + i\sin \theta)$. Always.

Now if $z = r(\cos\theta + i\sin \theta)= r\cos\theta + ri\sin\theta$ and $|z| = \sqrt{(r\cos\theta)^2 + (r\sin\theta)^2}= \sqrt{r^2(\cos^2\theta +\sin^2\theta) } =\sqrt{r^2\cdot 1} = r$.

(which we already knew! $r$ was defined to be $|z|$ and the beauty of polar coordinates is that $\cos^2\theta + \sin^2 \theta = 1$ so we never have to worry about calculating them.)

Okay but what is $|z^2|$?

Well $z^2 = [r(\cos\theta + i\sin\theta)]^2 = r^2 (\cos\theta + i\sin\theta)^2=$

$r^2[\cos^2\theta + 2\cos\theta\sin\theta i + i^2\cdot\sin^2 \theta)=$

$r^2[(\cos^2\theta - \sin^2\theta) + i\cdot 2\cos\theta \sin\theta]$.

Now by trig identities we know $\cos(2\theta) = \cos^2\theta - \sin^2\theta$ and $\sin(2\theta)$ so

$z^2 = r^2(\cos 2\theta + i \sin 2\theta)$.

And so $|z^2| = |r^2(\cos 2\theta + i \sin 2\theta)|$ but as above that is

$|r^2\cos 2\theta + ir^2\sin 2\theta|=$

$\sqrt{r^4(\cos^2 2\theta + \sin^2 2\theta}=\sqrt{r^4\cdot 1} = r^2 =|z|^2$.

And that's that.