Prove or disprove: complex numbers

Question

There is this question about complex numbers that I got stuck on for a while, I'd be glad for help:

Let $$z\in\Bbb{C}$$ such that $$z^3=\bar{z^3}$$ and $$|z|=1$$

Prove or disprove the followings: 1) there are exactly 6 different numbers that make the 2 equations true.

2) there are exactly 3 rational numbers that make the 2 equations true.

3) there exists infinity amount of different numbers that make the equation $$z^3=\bar{z^3}$$ true.

4) there exists exactly 4 numbers which are making the 2 equations true and their Imaginary part is bigger than 0.

How do I approach this kind of question? Again, I'd be glad for help.

Thanks in advance <3

Answer 1

HINT

We need to solve

$$z^3=\bar{z^3} \qquad |z|=1\iff e^{3i\theta}=e^{-3i\theta}\iff e^{6i\theta}=1$$

Answer 2

Multiply by $z^3$ and

$$z^6=z^3\bar z^3=(|z|^2)^3=1$$ so that the solutions are the sixth roots of one.

1) Yes.

2) No, 2.

3) Yes, if the modulus isn't constrained.

4) No, $3$.

Answer 3

Hint:

You can simply solve this equation and see if some answer fits what you've found.

It's better to use the exponential notation here: $|z|=1$ means $z=\mathrm e^{i\theta}\:$ for some $ 0\le \theta<2\pi$, and in this case $\bar z=z^{-1}$.

The equation $z^3=\bar z^{\mkern1mu3}$ translates as $$z^3=z^{-3}\iff z^6=\mathrm{e}^{6i\theta}=1\iff \theta\equiv 0\mod 2\pi, $$ i.e. $z$ is one of the sixth roots of unity.

Answer 4

Let $z=a+bi$ and $z*=a-bi$ ($z*$ is conjugate of $z$). We have 2 equations: $(a+bi)^3=(a-bi)^3$ and $a^2+b^2=1$. Expanding the first equation: $a^3+3bia^2 - 3ab^2- ib^3 = a^3 - 3bia^2-3ab^2 + ib^3$ Then by rearranging the terms and canceling we get: $3bia^2=ib^3$. Then: $3ba^2=b^3$. Now from the second equation we get: $a^2 = 1- b^2$.
By plugging in, that, in the first equation we get: $3b(1 - b^2)=b^3$.
One of the roots of this equation is $b=0$, so now we can divide the equation by $b$. Now we get: $3 - 3b^2 = b^2$
$4b^2 - 3 = 0$
$b = √3/2$ or $b = -√3/2$. Now for b = 0 we get a=1 or -1. For b = √3/2 we get a = 1/2 or -1/2. For b = -√3/2 we get a = 1/2 or -1/2. So we get 6 answers:$1;-1;1/2+√3/2i;1/2-√3/2i;-1/2+√3/2i;-1/2-√3/2i$ I hope my answer will help you. If something is not clear, feel free to ask.