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Combinatorial Proof Question

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.

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$$ z^3 = (a+ib)^3 \Rightarrow $$

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

$$ a^3 – b^2a – 2b^2a + i (2a^2b + b^2a – b^3) $$

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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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:

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

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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