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

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:


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?

  • 1
    $\begingroup$ 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.. $\endgroup$ Sep 14 '15 at 18:34
  • $\begingroup$ It should be $y=x-3$. $\endgroup$
    – mathlove
    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.


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