I'd like to solve for a complex number $z$ given, say, $$Re(az)=1,\quad Re(bz)=1$$ where $a$ and $b$ are known linearly independent (over $\mathbb R$) complex numbers. We can also assume $|a|=|b|=1$ if that matters.
Of course one can just solve a 2x2 linear system for the real and imaginary parts but I'm wondering if there's a slicker way to solve for $z$.
On
Without the $\,|a|=|b|=1\,$ assumption, $\,z\,$ can be determined by eliminating $\,\bar z\,$ between the two equations $\,a z + \bar a \bar z = b z + \bar b \bar z = 2\,$. The second one gives $\displaystyle \bar z = \frac{2-bz}{\bar b}\,$, then:
$$ a z + \bar a \frac{2-bz}{\bar b} = 2 \;\;\iff\;\; a \bar b z+\bar a(2-bz) = 2 \bar b \;\;\iff\;\; (a \bar b - \bar a b) z = 2(\bar b - \bar a) $$
If $|a| = |b| = 1$, $az$ and $bz$ both have absolute value $|z|$ and the same real part $1$, but are not equal. Therefore they must be complex conjugates: $a z = \overline{bz}$. Writing $z = r e^{i\theta}$, we have $z/\overline{z} = e^{2i\theta} = \overline{b}/a$. One of the two square roots of this is $e^{i\theta}$, and $r = 1/\cos(\theta)$.