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We have the following equation:

$$z^3 = \overline{z} $$

I set z to be $z = a + ib$ and since I know that $ \overline{z} = a - ib$. I was trying to solve it by opening the left side of the equation.

$$ z^3 = (a+ib)^3 \Rightarrow $$ $$ [a^2+b^2+i(ab + ba)](a+ib) \Rightarrow $$ $$ a^3 - b^2a - 2b^2a + i (2a^2b + b^2a - b^3) $$

but this is where I got so far and I'm not sure how continue and if my solution so far is even the right way to solve it.

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    $\begingroup$ Use $z=re^{i\theta}$ and $\bar z= re^{- i\theta}$ $\endgroup$
    – Someone
    May 15, 2015 at 9:26
  • $\begingroup$ @Mann by the 're' sign you mean the real number part? with a power of what? $\endgroup$
    – D_R
    May 15, 2015 at 9:28
  • $\begingroup$ It's the representation of complex number in polar coordinates. Where $r$ is magnitude of radius vector. $\theta$ tells the rotation. $\endgroup$
    – Someone
    May 15, 2015 at 9:29
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    $\begingroup$ You can use $\Rightarrow$ \Rightarrow$ or $\implies$ $\implies$. It definitely looks better than $=>$. However, in this case writing simply $=$ might be better. $\endgroup$
    – Martin Sleziak
    May 15, 2015 at 11:43

7 Answers 7

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In polar form, on the modulus side, $$r^3=r,$$ hence $r=0\lor r=1$.

On the argument side,

$$3\theta=-\theta+2k\pi,$$ hence $\theta=k\pi/2$.

The solutions are $$0,1,i,-1,-i.$$

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  • $\begingroup$ Why this downvote, please ? $\endgroup$
    – user65203
    May 15, 2015 at 9:30
  • $\begingroup$ A clear and simple answer. Why downvote? $\endgroup$
    – Empiricist
    May 15, 2015 at 9:31
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    $\begingroup$ Good answer, stupid downvote ! So $+1$. Cheers. $\endgroup$
    – Claude Leibovici
    May 15, 2015 at 9:33
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one of solutions is obviusly $z=0$. For other solutions the simple way is to write $z=re^{ia}$, then $$ z^3=\bar z\implies |z^3|=|\bar z|=|z|\implies |z|^2=1\implies r=1 $$ now we calculate $a$ by observing that $$ z^4=z\bar z=1 $$ so you just need to find all $4$-th roots of unity to finish...

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  • $\begingroup$ You never really used $re^{ia}$. You've just shown that $\lvert z\rvert^2=1$, and then noticed that $z^4=z\overline z=\lvert z\rvert^2=1$. $\endgroup$
    – Akiva Weinberger
    May 15, 2015 at 9:59
  • $\begingroup$ I noticed that for last step to find $a$... $\endgroup$
    – k1.M
    May 15, 2015 at 10:14
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Equating the real & the imaginary parts, $$a^3-3ab^2=a\iff a(a^2-3b^2-1)=0$$

and $$3a^2b-b^3=-b\iff b(3a^2-b^2+1)=0$$

Either $a=0\ \ \ \ (1)$ or $a^2-3b^2-1=0\ \ \ \ (2)$

and either $b=0\ \ \ \ (3)$ or $3a^2-b^2+1=0\ \ \ \ (4)$

Test with

$(1),(3);$

$(1),(4);$

$(2),(3);$

$(2),(4)$

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  • $\begingroup$ Hint: (2)-(4) yields $a^2=3b^2+1$ then $8b^2+4=0$ (!) $\endgroup$
    – user65203
    May 15, 2015 at 12:57
  • $\begingroup$ @YvesDaoust, That leads to absurdity as $a,b$ are real. $\endgroup$
    – lab bhattacharjee
    May 15, 2015 at 13:07
  • $\begingroup$ That's exactly the reason of the exclamation mark. $\endgroup$
    – user65203
    May 15, 2015 at 14:05
  • $\begingroup$ @YvesDaoust, $(2),(3)$ also leads to absurdity. But the rest two gives valid solutions, right? $\endgroup$
    – lab bhattacharjee
    May 16, 2015 at 5:35
  • $\begingroup$ Er, (2)-(3) leads to $a=\pm1$ I guess. $\endgroup$
    – user65203
    May 16, 2015 at 7:30
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Hint:

Multiplyng both sides by $z$ you have: $$ z^4=|z|^2 $$ Now use the polar form $z=|z|e^{i\theta}$.

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So good, so far.

Now: you want $z^3=\overline z$, therefore $\color{blue}{(a^3−3b^2a)}+i\color{green}{(2a^2b+b^2a−b^3)} =\color{blue}{a}-i\color{green}{b}$.

That gives you two equations with two unknowns. Solve them

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Multiplying both sides by $z$ (that does not introduce new solutions), $z^4=|z|^2$ is a real number.

Then by the imaginary part, $$4a^3b-4ab^3=0,$$ $$a=0\lor b=0\lor a^2=b^2.$$

The real part gives,

$$a^4-6a^2b^2+b^4=a^2+b^2$$ which simplifies to $$b^4=b^2\lor a^4=a^2\lor-4a^4=2a^2,$$ and it is an easy matter to list the five solutions,

$$\color{green}{(0,0),(1,0),(-1,0),(0,1),(0,-1)}.$$

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You have a good start. Rewrite equation as $z^3-\bar{z} =0$, now do $z=a+bi$, so we get $$a^3-3b^2a-a+i(3a^2b-b^3-b) = 0$$

Now both the imaginary and the real part must be equal to zero, so we get the following system of equations $$a^3-3b^2a-a=0 \wedge 3a^2b-b^3-b=0$$

Factoring gives:

$$a(a^2-3b^2-1)=0 \wedge b(3a^2-b^2-1)=0$$

So we have four possibilities:

  1. $a=0, b=0$
  2. $a=0, 3a^2-b^2-1=0$
  3. $a^2-3b^2-1=0, b=0$
  4. $a^2-3b^2-1=0, 3a^2-b^2-1=0$

First one clearly gives $z=0$.

Second one: Substitute $a=0$ in to get $b^2-1=0$, so $b=1$ or $b=-1$.

This gives $z=i$ and $z=-i$.

Third one: Substitute $b=0$ in to get $a^2-1=0$, so $a=1$ or $a=-1$.

This gives $z=1$ and $z=-1$.

Fourth one: Subtract the first equation trice form the second. This gives $8b^2+2=0$, so $b^2+\frac{1}{4}=0$, so $b=\pm\frac{1}{2}i$. This gives no solutions, since we defined $b = \Im(z)$ and it must be real.

Conclusion: The solutions are $z=0,1,-1,i,-i$.

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  • $\begingroup$ What happened to the solutions $b=\pm i/2$ ? $\endgroup$
    – user65203
    May 15, 2015 at 16:09
  • $\begingroup$ @YvesDaoust Note we have $b=1/2i$, so that also couts for the real part. So we have $-1/2-1/2=-1$, $-1/2+1/2=0$, $1/2+1/2=1$, $\endgroup$
    – wythagoras
    May 15, 2015 at 16:11
  • $\begingroup$ But what's $a$ and why do you add $\pm1/2\pm1/2$ ? $\endgroup$
    – user65203
    May 15, 2015 at 16:13
  • $\begingroup$ a=Re(z), b=Im(z). If b is not real, then the imaginary part moves to the real part. $\endgroup$
    – wythagoras
    May 15, 2015 at 16:16
  • $\begingroup$ Obviously there is a flaw as $\Im(z)$ is a real number and cannot be $\pm i/2$, but you don't answer the question: how do you get the value of $a$ and why do you combine $\pm1/2\pm1/2$ ? $\endgroup$
    – user65203
    May 15, 2015 at 16:20

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