Stuck solving $z^2$ + $zw^*$= $18$ and $2z^*$=$w^*(1−i)$ as a system of equations Let $z$ and $w$ be complex numbers that satisfy
$z^2$ + $z\overline{w}$= $18$
and $2\overline{z}$=$\overline{w}(1−i)$
with $\Re(z)>0$. Find $w$.
I tried to find $z$ by subbing $\overline{w}$ = $\frac{18-z^2}z$ into $2\overline{z}$=$\overline{w}(1−i)$ and letting $z\overline{z}=1$.
and got $z^2 = 17 + i$. This was where i got stuck. Using De Moivre's Theorem to get $z$ but the result was irrational.
 A: Preliminaries
$z = |z|e^{i\theta}\\
\bar z = |z|e^{-i\theta}\\
w = |w|e^{i\phi}\\
\bar w = |w|e^{-i\phi}\\
(1-i) = \sqrt 2 e^{-\frac {\pi}{4} i}$
Let's use the second equation to express $w$ in terms of $z.$
$2\bar z = \bar w(1-i)\\
2|z|e^{-i\theta} = (|w| e^{-i\phi})\sqrt 2 e^{-\frac {\pi}{4} i}\\
2|z|e^{-i\theta} = (\sqrt 2 |w|) e^{(-\phi}-\frac {\pi}4) i)$
Both the modulus and the argument are equal.
$|w|=\sqrt 2 |z|\\
\phi = \theta - \frac {\pi}{4}$
And substitute into the first equation.
$z^2 + z\bar w = 18\\
|z|^2 e^{2\theta i} + \sqrt 2 |z|^2 e^{\theta i - (\theta -\frac {\pi}{4})i} = 18\\
|z|^2 (e^{2\theta i} + \sqrt 2  e^{\frac {\pi}{4})i}) = 18\\
|z|^2 (\cos 2\theta + i\sin 2\theta + 1+i) = 18$
$|z|^2(\cos 2\theta + 1) = 18\\
|z|^2(\sin 2\theta + 1) = 0\\
\theta = -\frac{\pi}{4}\\
|z|^2 = 18$
$z = 3- 3i\\
w = -6 i$
Alternatively,
$z = x + yi\\
w = a + bi\\
2(x+yi) = (a-bi)(1-i)\\
2x + 2yi = a-b - (a+b) i\\
2x = a-b\\
2y = -(a+b)$
$(x^2 -y^2) + (2xy)i + (ax + by) + (ay-bx) = 18\\
x^2 - y^2 + ax + by = 18\\
2xy + ay - bx = 0$
$-a^2 + b^2 - a^2 - ab + ab - b^2 = 0\\
a^2 = 0\\
b^2 = 36$
A: Here is a slightly different approach:
Solve the second equation for $$\bar w = (1+i) \bar z$$ and substitute into the first to get $$z^2 + (1+i)z \bar z = 18.$$
Since $z$ is clearly nonzero, we can divide: $$\hat z^2 + (1 + i) = \frac{18}{|z|^2},$$ or $$\hat z^2 + i = \frac{18}{|z|^2} - 1.$$
Since the right-hand side is real, $\hat z^2$ is an element of the unit circle such that $\hat z^2 + i$ is real, forcing $$\hat z^2 + i = \frac{18}{|z|^2} - 1 = 0.$$ Since $z$ has positive real part, we have $$\hat z = \frac{1-i}{\sqrt 2}, |z| = 3 \sqrt 2,$$
and therefore $$z = 3(1 - i).$$ $$w = -6i.$$
