<s>[INST] Finding $z$ such that $\Re(\frac{z}{-3+2i}) = \frac{10}{13},\;|3-2\overline{z}| = |2z+3i|$ I wish to find $z$ such that $\displaystyle \Re(\frac{z}{-3+2i}) = \frac{10}{13},\;|3-2\overline{z}| = |2z+3i|$.
We may simplify $\displaystyle \frac{z}{-3+2i} = \frac{x+iy}{-3+2i} \frac{-3-2i}{-3-2i} = \frac{-3x+2y + i(-2x+3y)}{13}$.
Taking the real part of the fraction above gives us $\displaystyle \frac{-3x+2y}{13}$
Then, $-3x+2y = 10$
So now we move onto the other equation. 
Recall that $|z|=z\overline{z}$ and $z=x+iy$. Then, $|3-2(x-iy)| = |2(x+iy)+3i|$, $|3-2x+2iy| = |2x+2iy+3i|$,
Then, $$(3-2x+2iy)(3-2x-2iy) = (2x+i(3+2y))(2x-i(3+2y))$$ 
which simplifies to $$4x^2-12x+4y^2+9 = 4x^2 - i(2y+3)^2$$.
But this is still messy and does not simplify further. Can anyone see where I messed up?
 [/INST] again see last line $4x^2-12x+4y^2+9 = 4x^2 - i(2y+3)^2$, you have to wright
$4x^2-12x+4y^2+9 = 4x^2 +(2y+3)^2$
I sugesst you use complex modulus it will help you to reach answer in short way. $|z_1|=|z_2|$the $x_{2}^{2}+y_{2}^{2}=|z_1|^2=|z_2|^2=x_{1}^{2}+y_{1}^{2}$
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