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 quotientis 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
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