Solving complex numbers equation $z^3 = \overline{z} $

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.

Solutions Collecting From Web of "Solving complex numbers equation $z^3 = \overline{z} $"

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 $$0,1,i,-1,-i.$$

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…

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

$(1),(3);$

$(1),(4);$

$(2),(3);$

$(2),(4)$

Hint:

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

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

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,

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

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 $z=0,1,-1,i,-i$.