Complex numbers with trig

Question

I need help with this problem - Let $z = 2 e^{10 \pi i/21} $ and $w = e^{\pi i/7}$. Then if $(z-w)^6 = r e^{i\theta},$where $r \geq 0$ and $0 \leq \theta < 2\pi$, what is the ordered pair $(r, \theta)$?

I know that the magnitude is $\sqrt 3,$ but I can't seem to find the angle measure (in radians). Thank you in advance! An explanation would be appreciated.

Answer 1

$$z-w=2e^{10\pi i/21}-e^{\pi i/7}=e^{\pi i/7}\left(2e^{\pi i/3}-1\right)$$

But $\;e^{\pi i/3}=\frac12+\frac{\sqrt3}2i\implies 2e^{\pi i/3}-1=\sqrt3\,i\;$ , so

$$(z-w)^6=e^{6\pi i/7}\left(\sqrt3\,i\right)^6=-27e^{\pi i/7}$$

Answer 2

Consider $$ z-w=e^{\pi i/7}(2e^{\pi i/3}-1) $$ Next $$ 2e^{\pi i/3}-1=2\left(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}\right)-1=i\sqrt{3} $$ Therefore, as $\pi/2+\pi/7=9\pi/7$, $$ z-w=\sqrt{3}e^{9\pi i/7} $$ Can you finish?

Answer 3

Hint:

Let $O$ be the origin, $A$ and $B$ the affixes of $w$ and $z$ respectively.

Observe that $OA^2+AB^2=OB^2$, hence $OAB$ is a right triangle with hypotenuse $OB$.

Can you deduce $ \arg(z-w)$?