showing that $(z_1^2z_2)^2$ is a real number

Question

Given $z_1=a+bi,z_2=c+di,\frac{b}{a}=\frac{d}{c}=\frac{1}{\sqrt3}$,

$a,b,c,d$ are real numbers; $z_1,z_2$ are complex numbers.

Need to prove that $(z_1^2z_2)^2$ is a real number.

So i figured that $arg(z_1)=30^\circ,210^\circ,arg(z_2)=210^\circ,30^\circ$. but don't know how to proceed.

Answer 1

Hint

Write $b=\frac{a}{\sqrt{3}}$, $d=\frac{c}{\sqrt{3}}$. Now compute $z_1^2 \times z_2$. You will find something nice. Square the result.

I am sure that you can take from here.

Answer 2

you can get $\theta_1=arg(z_1)$ by using the fact that $\tan(arg(z_1))=\frac{1}{\sqrt{3}}$. Similarly you can find $\theta_2=arg(z_2)$. Then you can write $z_1=r_1e^{i\theta_1}$ and $z_2=r_2e^{i\theta_2}$. Then $(z_1^2z_2)^2=r_1^4r_2^2e^{i(4\theta_1+2\theta_2)}$. Now show that $4\theta_1+2\theta_2$ is some multiple of $\pi$, which is quite easy.

Answer 3

In any case $\arg(z_1^4)=120$ and $\arg(z_2^2)=60$.

Answer 4

Hint $\,\ z_1,\,z_2 = r i\zeta,\, s i\zeta,\,\ r,s\in\Bbb R,\,\ \color{#c00}{\zeta^3=1}\,\ $ so $\,\ (z_1^2 z_2)^2 = (-r^2s\, i\,\color{#c00}{\zeta^3})^2 \in\Bbb R.\ $ Explicitly

$$\begin{eqnarray} &&z_2 = (\sqrt{3}+i)d = -2di\left(\dfrac{-1-\sqrt{-3}}2\right) = -2di\zeta \\ && z_1 = (\sqrt{3}+i)b = -2bi\zeta\end{eqnarray}$$