Exercise: Describe geometrically, the set of all complex numbers z which satisfy the following condition
Since |z-1| >0, we know the set we want to describe does not contain the point z=1. By the triangle inequality, we have
for all z. So we want to exclude all points from the plane where the equality
holds. That is, we want to exclude any z whose distance from 1 is equal to 1 plus its distance to the origin. This just means we have to exclude the negative real axis and the origin. (Draw a picture.)
We can also see this algebraically. Writing z= x+iy we have
Setting these equal gives
which reduces to
This can only hold if and y=0.
So the set we want is the complex plane with the point z=1 and the segment deleted.
Do you need more help? Please post your question on our S.O.S. Mathematics CyberBoard.
Author: Michael O'Neill