Determine the set of complex numbers $z$ such that $z^6$ is a negative real

Question

I want to solve it using the trigonometric (and not the polar) form of complex numbers.

Let $z^6=-a$, being $a$ a positive real number.

$|-a|=a$ and the $-a$ argument is $π$ because $-a$ is a negative real number.

Then the six roots of $-a$ are:

$z_0=(a)^{1/6}(\cos(π/6)+i\sin(π/6))$

$z_1=(a)^{1/6}(\cos(π/2)+i\sin(π/2))$

$z_2=(a)^{1/6}(\cos(7π/6)+i\sin(7π/6))$

$z_3=(a)^{1/6}(\cos(3π/2)+i\sin(3π/2))$

$z_4=(a)^{1/6}(\cos(11π/6)+i\sin(11π/6))$

$z_5=(a)^{1/6}(\cos(5π/6)+i\sin(5π/6))$

These complex numbers $z_i$ with $0≤i<6$ ($i$ is a natural number) satisfy $z^6$ is a negative real number but these complex numbers are infinite because they depend from $a$ and $a$ is any positive real number, that is, the values they can take are infinite so how do I specify the set of complex numbers requested?

Answer 1

You're going backwards; the direct approach is easier.

If you write $z=a(\cos\alpha+i\sin\alpha)$, then $$ z^6=a^6(\cos6\alpha+i\sin6\alpha) $$ (De Moivre), which is a negative real if and only if $\sin6\alpha=0$ and $\cos6\alpha<0$.

Can you go on? Yes, $a$ can be any positive real, but $\alpha$ is restricted.