Question

How to prove perpendicular lines in complex plane?
Question: Let w = a+ bi not 0. Use Euclidean geometry to explain why the set {z in C : |z-w| =|z+w|} is a straight line through the origin 0 perpendicular to the line through -w and w. Then verify the result analytically.
I am done with part a and most of part b until Re (z w(bar)) =0, then stuck
Please, help

Answer 1

Here is my work:
|z -w| = |z +w| \(\implies |z-w|^2 =|z+w|^2\)
Expand and simplify, I got \(4Re z\bar w=0\implies Re z\bar w=0\), then???

Answer 2

Let \(z=x+iy\)

the required line is \(ax+by=0\)

Answer 3

the line through \(-w\) and \(w\) has a direction vector \((a, b)\)

Answer 4

the line \(ax+by=0\) has a direction vector \((-b,a)\)

Answer 5

the dot product of direction vectors is \(0\), therefore the lines are perpendicular

Answer 6

I am sorry. I didn't see your reply.
Thanks for the explanation.
Question: how can we prove it passes through origin?

Answer 7

\(ax+by=0\) satisfies the point \((0,0)\)