I'm trying to algebraically find the equation of the perpendicular bisector of line joining (-3+2i) (3-2i). I'm Trying to use the identity z x z*=|z|^2 but u keep getting the solution x+iy=0 Instead of an equation. Can anyone pls point out my error
For reference, I am using this method From the equation $|z-a|=|z-b|$ in the complex plane, obtain the slope-intercept equation of the line
The equation to solve is $$ |z+3-2i|=|z-3+2i| $$ Squaring and using $|w|^2=w\bar{w}$, you get $$ (z+3-2i)(\bar{z}+3+2i)=(z-3+2i)(\bar{z}-3-2i) $$ This yields $$ z\bar{z}+(3-2i)\bar{z}+(3+2i)z+13=z\bar{z}+(-3+2i)\bar{z}+(-3-2i)z+13 $$ hence $$ (3+2i+3+2i)z+(3-2i+3-2i)\bar{z}=0 $$ that becomes $$ (6+4i)z+(6-4i)\bar{z}=0 $$ This means that the number $(6+4i)z$ is purely imaginary, so you can express the solutions as $$ z=\dfrac{26ki}{6+4i}=\frac{13ki}{3+2i}=ki(3+2i)=k(-2+3i) $$ with real $k$. In standard coordinates this is the line $3x-2y=0$.
Where is your error? You get $$ 3(z+\bar{z})+2(z-\bar{z})i=0 $$ but this isn't a representation of a complex number as real part plus $i$ times the imaginary part: indeed $z-\bar{z}$ is a purely imaginary number, so $(z-\bar{z})i$ is real.
If you write $z=x+yi$, then the relation becomes $$ 6x+2(2yi)i=0 $$ hence $6x-4y=0$, which is the same as $3x-2y=0$ as found above.