This may seem like a really easy question to you guys, but I'm getting confused.
So I know that z and w lie on the unit circle cause they have a modulus of 1.
And I've turned z$\overline{w}$ into polar form which gave me 1cis$\frac{\pi}{3}$ so I know that that lies on unit circle too (obviously anyway cause z and w lie on it anyway)
From what I know when multiplying complex numbers I can just add their $\theta$, so I need two that add to $\frac{\pi}{3}$ which is quite a few of numbers.
I tested (1cis$\frac{\pi}{6}$)$^{2}$ which should give me 1cis$\frac{\pi}{3}$, and in standard for this is just ($\frac{\sqrt 3}{2} + \frac{1}{2}i$)$^{2}$ but as I'm working with conjugate I would do
($\frac{\sqrt 3}{2} + \frac{1}{2}i$) - ($\frac{\sqrt 3}{2} - \frac{1}{2}i$) = i which will have a modulus of 1, even when squared.
But then I also tested (1cis$\frac{4\pi}{3}$)($\pi$) again should give me 1cis$\frac{\pi}{3}$, which again as I'm working with conjugate I would do
(-$\frac{1}{2}$-$\frac{\sqrt 3}{2}i$)(-1+0i) = $\frac{1}{2}$-$\frac{\sqrt 3}{2}i$ which will have a modulus of 2 and then squaring it will give me 4.
After testing those two I just got really confused about if the question has just 1 answer. Any insight would help thanks.
Hint: For any two complex numbers $z$ and $w$, can $|z-w|^2$ be expressed in terms of just $|z|$, $|w|$, and $z \bar w$ in a straightforward manner? Indeed, $$|z - w|^2 = (z - w)(\bar z - \bar w) = z \bar z - (w \bar z + z \bar w) + w \bar w$$ Do you see $|z|$ and $|w|$ lurking in this expression? And what is the relation between $z \bar w$ and $w \bar z$?