Finding complex number defined by 3 equations

Question

Let $z$ be a complex number satisfying

$$\DeclareMathOperator{\Re}{Re}\Re[z^4]=1/2$$

$$z\bar{z}+2|z|-3=0$$

$$\arg z \leq \frac{\pi}{4}.$$

Answer 1

Your second equation says, $\vert z \vert ^{2}+2\vert z\vert -3=0$ so $(|z|+3)(\vert z \vert-1)=0$ since $|z|+3>0$ we must have $|z|=1$, hence $z=e^{i\theta}$ for some $\theta \in [0,2 \pi[$. $z^{4}=e^{i4\theta}$. Re$(z^{4})=\cos(4\theta)$, hence $\cos 4 \theta=\frac{1}{2}$ so $4 \theta=\frac{\pi}{3}$ or $4 \theta=\frac{5\pi}{3}$ hence $\theta=\frac{\pi}{12}$ or $\theta=\frac{5 \pi}{12}$, but only $\theta=\frac{\pi}{12}$ is smaller than $\frac{\pi}{4}$. So $z=e^{i\frac{\pi}{12}}$ (Edited thanks to comments).

Answer 2

Like with many things complex, this is best looked at graphically.

The first equation defines some lines. The second equation defines a circle. The third equation defines a wedge.

Look at the intersection.

Answer 3

If $ x + i y = z$

then you have

$$ x^4 - 6 x^2 y^2 + y^4 = \frac 12$$

Let $$ x = r \cos \theta , \,\, y = r \sin \theta $$

Then

$$ 1- 8 (\cos \theta \,\cos \theta )^2 = 1/2 $$

$$ \theta = \pi/12, \pi/2 - \pi/12 $$

$$ r^2 +2 r - 3 =0, r =1, r= -3 $$