Find all complex numbers $z$ such that $|z| + \overline{z}=9-3i$.
Thinking about using with the fact that $z=a+bi$ and $|z| = \sqrt{a^2+b^2}$ in some way. Not sure how.
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$\begingroup$Write $\;z=x+iy\;$ , so that your equation is
$$\sqrt{x^2+y^2}+x-iy=9-3i\stackrel{\text{compare real and imag. parts}}\implies\begin{cases}\sqrt{x^2+y^2}+x=9\\{}\\y=3\end{cases}\;\;\implies$$
$$\sqrt{x^2+3^2}+x=9\implies x^2+9=81-18x+x^2\implies18x=72\implies x=4$$
and the unique solution is $\;4+3i\;$
$\endgroup$ $\begingroup$$\overline{z}=-|z|+9-3i\,$, then taking the magnitudes of the two sides:
$$\require{cancel} |z|^2=(-|z|+9)^2+3^2 \;\;\iff\;\; \cancel{|z|^2}=\cancel{|z|^2}-18 |z| +90 \;\;\iff\;\; |z|=5 $$
Substitute $|z|=5$ back into the given equation, then $\bar z = 4-3i \iff z = 4+3i\,$.
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