if z1=10+6i and z2=4+6i if z is any complex no such that the argument of (z-z1)/(z-z2)
is pi/4 then prove that mod(z-7-9i)=3sqrt(2)
can yu solve it graphically
Graphically, the argument of the quotient is the angle between the lines and . By the theorem about angles in the same segment (or rather its converse), that says that z is on a circle through and . Another theorem (the one about the angle at the centre C being twice the angle at the circumference) says that the lines and must be at right angles. It's easy to deduce from that (see the diagram) that C must be at the point 7+9i and that the radius must be . So z lies on the circle