$z^3=w^3 \implies z=w$? I've reached this in another problem I have to solve: $z,w \in \Bbb {C}$.  $z^3=w^3 \implies z=w$? 
I've scratched my head quite a bit, but I completely forgot how to do this, I don't know if this is correct:
$$
z^3=|z^3|e^{3ix}=|w^3|e^{3iy}
$$
I know the absolute values are equal, so I get $3ix=3iy \implies x=y \implies z=w$. I'm not sure I solved this correctly, I know this is pretty basic, but I haven't done this stuff in a year...
 A: Not true. For example, $z = 1 \neq w = e^{2\pi i/3}$ but $z^3 = w^3 = 1$.
A: Take a cube root of unity other than $1$, and call it $w$. Thus $w \neq 1$, and $w^3 = 1$. Take another complex number, $z = i$, then clearly $i \neq iw$, but $i^3 = 1\cdot i^3 = w^3\cdot i^3 = (i\cdot w)^3$
A: We have, $$z^3=\omega^3\iff z^3-\omega^3=0$$ Since above equation is cubic in terms of $z$ hence, using $a^3-b^3=(a-b)(a^2+b^2+ab)$ we have $$(z-\omega)(z^2+\omega^2+\omega z)=0$$ $$\implies z-\omega=0 $$$$\implies \color{blue}{z=\omega}$$ $$\implies z^2+\omega^2+\omega z=0$$ Solving above quadratic equation for $z$ we gat $$ z=\frac{-\omega\pm\sqrt{\omega^2-4(1)(\omega^2)}}{2\cdot 1}$$ $$=\frac{-\omega\pm\sqrt{-3\omega^2}}{2}=\frac{-\omega\pm i\omega\sqrt{3}}{2}$$ Hence, there are three roots of the given cubic equation given as follows $$\bbox[5px, border:2px solid #C0A000]{\color{red}{ z=\omega,\ \frac{-\omega+ i\omega\sqrt{3}}{2},\ \frac{-\omega- i\omega\sqrt{3}}{2} }}$$
A: Presumably you can already compute the three cube roots of -1 (say).
Can you see that each of them have the same cube (-1 in this case)? Are these cube roots equal?
Hopefully from there you can see how to answer your question.
A: We assume $z \ne 0 \ne w$; it is obvious that in the other circumstance $z = w = 0$ is the only possible solution.
Now,  since
$z^3 = w^3 \Leftrightarrow z^3 - w^3 = 0, \tag{1}$
then
$(z - w)(z^2 + zw + w^2) =  z^3 - w^3 = 0; \tag{2}$
we see that if
$z^2 + zw + w^2 \ne 0, \tag{3}$
then
$z = w; \tag{4}$
furthermore, if (4) binds then
$z^2 + zw + w^2 = 3z^2 \ne 0; \tag{5}$
thus
$z = w \Leftrightarrow z^2 + zw + w^2 \ne 0; \tag{6}$
certainly $z = w$ is a possible solution to (1); (6) indicates that the any other prospects are to be found via the logically equivalent 
$z \ne w \Leftrightarrow z^2 + zw + w^2 = 0; \tag{7}$
we therefore scrutinize
$z^2 + zw + w^2 = 0 \tag{8}$
for solutions of (1) other than $z = w$.  Perhaps the most straightforward and clearest way to proceed is to exploit the assumption $w \ne 0$ and divide (8) through by $w^2$:
$(\dfrac{z}{w})^2 + \dfrac{z}{w} + 1 = 0; \tag{9}$
if we now set
$\omega = \dfrac{z}{w}, \tag{10}$
we find
$\omega^2 + \omega + 1 = 0. \tag{11}$
(11) is readily recognized as the equation for the non-real cube roots of unity; indeed we have, taking for the moment $z = \omega$ and $w = 1$ in (2),
$(\omega - 1)(\omega^2 + \omega + 1) = \omega^3 - 1 = 0, \tag{12}$
which shows that if
$\omega^3 = 1, \;\; \omega \ne 1, \tag{13}$
then $\omega$ satisfies (11);  it is easily seen from the quadratic formula that the two such $\omega$ are
$\omega = -\dfrac{1}{2} \pm \dfrac{i}{2}\sqrt{3}; \tag{14}$
we recognize that
$\cos \dfrac{2\pi}{3} = \cos -\dfrac{2\pi}{3} = -\dfrac{1}{2},$
$\sin \dfrac{2\pi}{3} = \dfrac{\sqrt{3}}{2}; \tag{15}$
therefore, if $\omega \ne 1$ we have
$\omega = \cos \dfrac{2\pi}{3} \pm i \sin \dfrac{2\pi}{3}$
$=  e^{\pm 2\pi i /3} = e^{2\pi i / 3}, e^{4\pi i / 3}; \tag{16}$
now since
$\dfrac{z}{w} = \omega = e^{\pm 2\pi i / 3}, \tag{17}$
it follows that, other than $z = w$, the solutions to (1) are
$z = e^{\pm 2 \pi i /3} w.  \tag{18}$
(18) is easily verified:
$z^3 = (e^{\pm 2 \pi i / 3}w)^3 = w^3(e^{\pm 2\pi i / 3})^3$
$= w^3 e^{\pm 2 \pi i} = w^3, \tag{19}$
since $e^{\pm 2 \pi i} = 1$.  We thus see that (1) entails precisely three possibilities:
$z = w, z = e^{2\pi i /3} w, z = e^{- 2 \pi i /3} = e^{4 \pi i /3} w; \tag{20}$
it's really all about equations (11)-(12); and of course, there are the nearly "self-evident truths" of the  generalization of these results to the case $z^n = w^n$ for $n \in \Bbb Z$.  
