# Modulus of complex numbers loci problem

Given:

$|z-4+3i|$ = $|z-i|$

I need to describe and draw the locus. The work I have done so far is I converted the sides to their cartesian equivalents such as:

$|z-4+3i|^2$ = $|z-i|^2$

$(x-4)^2+(y+3)^2 = x^2 + (y-1)^2$

Which simplifies to:

$y=4x-15/2$

Which is a straight line. What I am not getting is how can two sides of modulus of complex numbers simplify into a line? Is what I am doing correct at all?

• Also, the given statement actually is the mathematical representation of "$z$ is equidistant from the points $(4, -3)$ and $(0,1)$ plotted on the complex plane where x-axis represents the real part and y-axis represents the imaginary part of the number" .. The locus, elementary geometry would tell us, is the perpendicular bisector of the segment that joins these two points.. Sep 14 '15 at 18:34
• It should be $y=x-3$. Sep 14 '15 at 19:16

Notice, we have $$|z-4+3i|=|z-i|$$ $$|z-(4-3i)|=|z-(0+i)|$$ The above equation shows that the distances of the point $z(x, y)$ from the two fixed points $(4, -3)$ & $(0, 1)$ are equal. Hence, the point $(x, y)$ will always lie on the perpendicular bisector of the line joining the points $(4, -3)$ & $(0, 1)$ in the complex plane.
The locus of the point $(x, y)$ is a straight line. Your answer is correct.