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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...

YoTengoUnLCD
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  • $e^a=e^b$ does not imply $a=b$, because $e^z$ is not one-to-one. Via de Moivre's formula, $e^z=1$ with $z=x+iy$ implies $x=0$ and $y=2\pi n$ for some integer $n$. Thus, $e^a=e^b$ implies $e^{a-b}=1$ implies $a-b=2\pi in$ for some integer $n$. Also, the plural of modulus is moduli. (Although "moduli" already has a cool technical meaning in advanced mathematics.) – anon Aug 04 '15 at 02:06
  • Thanks for the explanation! About the "modulus", I meant absolute value (modulus is very close to its spanish counterpart :) ). – YoTengoUnLCD Aug 05 '15 at 00:15
  • Bro, do you even do the roots of unity? – Alec Teal Aug 05 '15 at 00:24
  • I tried to, last night. I understand my question was pretty basic, but I had forgotten how to solve simple stuff like that. – YoTengoUnLCD Aug 05 '15 at 00:26
  • "Modulus" is also a term for "absolute value of a complex number" in English, it's just not as common as "absolute value" (and usually not used in the plural either). – anon Aug 05 '15 at 00:52

5 Answers5

6

Not true. For example, $z = 1 \neq w = e^{2\pi i/3}$ but $z^3 = w^3 = 1$.

Simon S
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  • Can you explain what's the mistake in my process taking me to that false result? – YoTengoUnLCD Aug 04 '15 at 00:42
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    When equating the arguments we don't have $3ix = 3iy$ but rather

    $$3ix = 3iy + 2k\pi i \quad\text{ for some integer } k$$

    and thus $$x = y + 2k\pi/3 \quad\text{ for some integer } k$$

     – Simon S Aug 04 '15 at 00:44
  • The argument are not necessarily equal. They should differ by an integer multiple of $2\pi$. – chhro Aug 04 '15 at 00:45
  • Oh, I remember that, thanks! What was the justification to add the $2k\pi$ tho? – YoTengoUnLCD Aug 04 '15 at 00:52
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    $$e^{\theta i + 2k\pi i} = e^{\theta i}e^{2k\pi i} = e^{\theta i} (e^{2\pi i})^k = e^{\theta i} \cdot 1^k = e^{\theta i}$$

    Hence if $e^{\theta i} = e^{\phi i}$, the most we can say is that for some integer $k$ we have $\theta = \phi + 2k\pi$.

     – Simon S Aug 04 '15 at 00:59
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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 $i^3 = 1\cdot i^3 = w^3\cdot i^3 = (i\cdot w)^3$

DeepSea
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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 $$\bbox[5px, border:2px solid #C0A000]{\color{red}{ z=\omega,\ \frac{-\omega+ i\omega\sqrt{3}}{2},\ \frac{-\omega- i\omega\sqrt{3}}{2} }}$$

Harish Chandra Rajpoot
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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.

Glen_b
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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 $n \in \Bbb Z$.

Robert Lewis
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