<s>[INST] Solving for $z$ given 3 constraints: $\DeclareMathOperator{\Re}{Re}\Re[z^4]=1/2$ , $z\bar{z}+2|z|-3=0$, $\arg z \leq \frac{\pi}{4}.$ 

Let $z$ be a complex number satisfying

$$\DeclareMathOperator{\Re}{Re}\Re[z^4]=\frac{1}{2}$$
$$z\bar{z}+2|z|-3=0$$
$$\arg z \leq \frac{\pi}{4}.$$
Find $z$

As a side note this is the same as this question here: Finding complex number defined by 3 equations , but I'm looking for a way to finish it using my approach.

I'm going to let $z = x+iy$ and attempt to solve for $x$ and $y$
$$\DeclareMathOperator{\Re}{Re}\Re[z^4]=\frac{1}{2}$$
$$\DeclareMathOperator{\Re}{Re}\Re[(x+iy)^4]=\frac{1}{2}$$
$$\DeclareMathOperator{\Re}{Re}\Re[x^4+y^4-6x^2y^2 +i(4x^3y-4xy^3)]=\frac{1}{2}$$
Considering real only
$$\DeclareMathOperator{\Re}{Re}\Re[x^4+y^4-6x^2y^2]=\frac{1}{2}$$
$$ x^4+y^4-6x^2y^2 = \frac{1}{2} $$

Now with the second equation
$$z\bar{z}+2|z|-3=0$$
$$ (x+iy)(x-iy) + 2\sqrt{x^2+y^2} - 3 = 0 $$
$$ x^2 + y^2 + 2\sqrt{x^2+y^2} - 3 = 0 $$

I will worry about the argument after I solve for $x$ and $y$.

I now have $2$ equations:
$$ x^4+y^4-6x^2y^2 = \frac{1}{2} $$
$$ x^2 + y^2 + 2\sqrt{x^2+y^2} - 3 = 0 $$
Working on the second one
$$ x^2 + y^2 + 2\sqrt{x^2+y^2} - 3 = 0 $$
Let $x^2+y^2$ $=$ $p$
$$ p + 2\sqrt{p} - 3 = 0 $$
$$ (\sqrt{p}-1)(\sqrt{p}+3) = 0 $$
$$ \sqrt{p} = 1 , \sqrt{p} \neq -3 $$
$$ x^2+y^2 =1 $$
Squaring both sides
$$ x^4 + y^4 + 2x^2y^2 = 1 $$
$$ x^4 + y^4 = 1-2x^2y^2 $$
Subbing this into the first equation
$$ x^4+y^4-6x^2y^2 = \frac{1}{2} $$
$$ 1-2x^2y^2-6x^2y^2 = \frac{1}{2} $$
$$ \frac{1}{2} = 8x^2y^2 $$
$$ \frac{1}{16} = x^2y^2 $$
$$ y^2 = \frac{1}{16x^2} $$
Subbing this into
$$ x^2+\frac{1}{16x^2} =1 $$
$$ x^4 - x^2 + \frac{1}{16} = 0 $$
$$ x^2 = \frac{2+\sqrt{3}}{4}  , x^2 = \frac{2-\sqrt{3}}{4}$$
$$ x = ± \frac{\sqrt{6}+\sqrt{2}}{4} , x= ± \frac{\sqrt{6}-\sqrt{2}}{4} $$
$$ y = ± \frac{\sqrt{6}-\sqrt{2}}{4} , y= ± \frac{\sqrt{6}+\sqrt{2}}{4}$$
Now I am stuck which set is correct to represent $z=x+iy$?
 [/INST] The third condition gives you - Both $x$ and $y$ are positive and $x>y$
So, from your set of answers(assuming they are right), the answer would be 
$x=\frac{\sqrt 6+\sqrt 2}{4}$
$y=\frac{\sqrt 6-\sqrt 2}{4}$
</s>