Complex number quadratic with conjugation

Question

I have the following equation: $$ 3z^2+z^*-1=0 $$ z is a complex number and $z^*$ is its conjugate. I'm stuck solving it. There's what I did so far: $$ z = a+bi $$ $$ 3a^2+6abi-3b^2+a+bi=1 $$ Assuming $Im(z) = 0$ I'm getting following equations: $$ \begin{cases} 3a^2-3b^2+a=1\\ 6abi-bi=0 \end{cases} $$ $$ bi(6a-1)=0=>b=0 \lor a=-\frac{1}{6} $$

Since $b=0$, I'm getting $$ 3a^2+a=1 $$ which will have 2 solutions. I have no idea what should I do next since I will be getting 3 different $a$'s, but wolfram shows me only 2 solutions. $$ z = -\frac{1}{6} - \frac{\sqrt{13}}{6} $$ $$ z = \frac{\sqrt{13}}{6} -\frac{1}{6} $$

Answer 1

If $a=-\dfrac16$, then $3a^2-3b^2+a=\dfrac1{12}-3b^2-\dfrac16=1\iff 3b^2=\dfrac1{12}-\dfrac16-1$,

which has no real solutions for $b$, so there really are only the two solutions from when $b=0$.

Answer 2

Well, we are trying to solve:

$$3\text{z}^2+\overline{\text{z}}-1=0\tag1$$

As you did, let's set $\text{z}:=\alpha+\beta i$ where $\alpha\in\mathbb{R}$ and $\beta\in\mathbb{R}$. So we get:

$$3\left(\alpha+\beta i\right)^2+\alpha-\beta i-1=0\tag2$$

Expanding $\left(\alpha+\beta i\right)^2$, gives:

$$\left(\alpha+\beta i\right)^2=\alpha^2-\beta^2+2\alpha\beta i\tag3$$

So, we get:

$$3\left(\alpha^2-\beta^2+2\alpha\beta i\right)+\alpha-\beta i-1=0\tag4$$

Finding the real and imaginary parts, we can set:

$$ \begin{cases} \Re\left(3\left(\alpha^2-\beta^2+2\alpha\beta i\right)+\alpha-\beta i-1\right)=0\\ \\ \Im\left(3\left(\alpha^2-\beta^2+2\alpha\beta i\right)+\alpha-\beta i-1\right)=0 \end{cases}\tag5 $$

This gives:

$$ \begin{cases} 3\alpha^2-3\beta^2+\alpha-1=0\\ \\ 6\alpha\beta-\beta=0 \end{cases}\tag6 $$

Solving the second equation for $\beta$ gives:

$$\beta=0\tag7$$

So:

$$3\alpha^2-3\cdot0^2+\alpha-1=0\space\Longleftrightarrow\space\alpha=\frac{\pm\sqrt{13}-1}{6}\tag8$$

As pointed out by @Gae. S., try to prove or disprove that $\alpha=\frac{1}{6}$ is or isn't a solution to your problem.

Answer 3

The given equation implies

$$ z=1-3(z^*)^2=1-3(1-3z^2)^2=-2+18z^2-27z^4 $$ which factorizes as $$(3z^2+z-1)(3z+1)(3z-2)=0$$ and yields the solutions $z=-\frac16\pm \frac{\sqrt{13}}6$, while $z=\frac23,-\frac13$ are invalid.