Cube root of a unit vector is also unit

Question

How do I prove or disprove this?

Let $z\in\mathbb C$. If $|z^3|=1$ then $|z|=1$.

My intuition tells me this isn't true, but I couldn't find an opposing example for this. I'd be glad for help.

Answer 1

Note that

$$|z^3|=|z|^3=1 \iff |z|=1$$

Let prove the more general $$|z^n|=|z|^n$$

which easily follows by exponential form

$$z=re^{ix}\implies z^n=r^ne^{nix}$$

since by definition $r>0$ we have

$$|z^n|=r^n=|z|^n\quad \square$$

As an alternative, by polar form, we can prove by trigonometric identities that

$$|z_1z_2|=|z_1||z_2|$$

$$z_1=r_1(\cos{x}+i\sin x)\quad z_2=r_2(\cos{y}+i\sin y)\implies z_1z_2=r_1r_2(\cos{(x+y)}+i\sin (x+y))$$

and from here $|z^n|=|z|^n$ easily follows.

Answer 2

If $|z^3|=1$, then $|z|^3=1$ and there is only one real number whose cube is $1$, which is $1$. So $|z|=1$.

Note that $|z^3|=|z|^3$ because $(\forall z,w\in\mathbb{C}):|zw|=|z|.|w|$.

Answer 3

$1=|z^3|=|z|^3$, hence $1=|z|$.

Answer 4

$z \in \mathbb{C}:$

$|z^3| = |z|^3 =1.$

Note that $|z| \in \mathbb{R^+}$.

$|z|^3=1$ , $|z|^3-1=0$, or

$(|z|-1)(|z|^2+|z|+1)=0$.

Hence : $|z|=1$, the only real solution.

Note:$ |z|^2 +|z| +1 >0$, since

$|z|^2+|z|+1=$

$(|z|+1/2)^2 -1/4+1 =$

$ (|z|+1/2)^2 +3/4 >0.$