How to solve $\arg z^3 =0$ where $z\in\mathbb{C}$?

Question

Can someone show me how to solve $$\arg z^3 = 0$$ I don't know even how to start because of the zero.

Answer 1

Hint:

First prove that given non-zero complex numbers $z,w$ that Arg$(zw)$ = Arg$(z) +~$ Arg$(w),~$ within a modulus of $(2\pi)$.

You will need formulas for $\cos(a + b)$ and $\sin(a + b).$

Then prove that for non-zero complex $z,~$ Arg$(z^3)$ = $3 \times $ Arg$(z),~$ within a modulus of $(2\pi)$.

Answer 2

A complex number $z \neq 0$ can be always rewritten as:

$$z = M e^{i\theta},$$

with $M > 0$ and $\theta \in [0, 2\pi]$.

Recall that $\theta \in [0, 2\pi]$, since $e^{i\theta + 2k\pi} = e^{i\theta}e^{i2k\pi} = e^{i\theta}\cdot 1 = e^{i\theta}$ for any integer $k$.

Moreover, $\text{Arg}(z) = \theta$ by definition.

Now, observe that:

$$z^3 = M^3 e^{i3\theta} = M^3 e^{i \phi},$$

since you can always find an integer $k$ such that $3 \theta = 2k \pi + \phi$ and $\phi \in [0, 2\pi].$ Then:

$$\text{Arg}(z^3) = \phi \equiv 3 \theta ~\text{mod}~2\pi.$$

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What does $\phi \equiv 3 \theta ~\text{mod}~2\pi$ mean?

It means that you can always find an integer $k \in \{\ldots, -2, -1, 0, 1, 2 \ldots\}$ such that:

$$3\theta = 2k\pi + \phi.$$

Example.

If $\theta = \frac{5}{6}\pi$, then:

$$3 \theta = \frac{5}{2}\pi = \left(2 + \frac{1}{2}\right)\pi = 2\pi + \frac{1}{2}\pi = 2k\pi + \phi,$$

where $k=1$ and $\phi = \frac{1}{2}\pi.$