I need help with the algebraic portion of this complex trig question

Question

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 have the starting idea of drawing it out on the graph table. Currently, I need some help with the algebraic portion of the problem.

Answer 1

$$(z-w)^6 = r e^{i\theta}$$ $$|(z-w)^6 |= |r e^{i\theta}|$$ $$|(z-w) |^6= r $$ $$z -w= e^{\pi i/7}(2e^{\pi i/3}-1)$$ $$|z -w|= |e^{\pi i/7}(2e^{\pi i/3}-1)|=|e^{\pi i/7}|\cdot|(2e^{\pi i/3}-1)|$$ $$|z -w|=1\cdot|(2e^{\pi i/3}-1)|=\sqrt{3}$$ $$r=|(z-w) |^6=27 $$

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$$Arg(z-w)^6=6\cdot Arg(z-w)$$ $$z-w=e^{\pi i/7}(2e^{\pi i/3}-1)$$ $$arg(z-w)=\frac{9\pi}{14}$$

Answer 2

HINT

We have that

• $r=|z-w|^6$

• $Arg((z-w)^6)=6\cdot Arg(z-w)$

• $Arg(z_1z_2)=Arg(z_1)+Arg(z_2)$

and

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