TITLE: Finding $z$ such that $\Re(\frac{z}{-3+2i}) = \frac{10}{13},\;|3-2\overline{z}| = |2z+3i|$
QUESTION [2 upvotes]: 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?

REPLY [0 votes]: 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 $x_{2}^{2}+y_{2}^{2}=|z_1|^2=|z_2|^2=x_{1}^{2}+y_{1}^{2}$

REPLY [0 votes]: Given $$|3-2\bar{z}| = |2z+3i|\Rightarrow |(2\bar{z}-3)| = |2z+3i|\;,$$ Now put $$z=x+iy$$
We get $$\displaystyle |(2x-3)+i(2y)| = |2x+i(2y+3)|\Rightarrow (2x-3)^2+4y^2=(2x)^2+(2y+3)^2$$
So we get $$\displaystyle 9-12x = 9+12y\Rightarrow x+y=0.....(1)$$
and First part $$\displaystyle \bf{\Re\left(\frac{z}{-3+2i}\right)} = \frac{10}{13},$$ You have got $$-3x+2y=10..............(2)$$
so we get $$\displaystyle x=-2,y=2$$, So we get $$z=x+iy=-2+2i$$
{"set_name": "stack_exchange", "score": 2, "question_id": 1436776}