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 $\displaystyle (z-z_1)/(z-z_2)$ is the angle between the lines $\displaystyle z\to z_1$ and $\displaystyle z\to z_2$. By the theorem about angles in the same segment (or rather its converse), that says that z is on a circle through $\displaystyle z_1$ and $\displaystyle z_2$. Another theorem (the one about the angle at the centre C being twice the angle at the circumference) says that the lines $\displaystyle z_1\to C$ and $\displaystyle z_2\to C$ 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 $\displaystyle 3\sqrt2$. So z lies on the circle $\displaystyle |z-(7+9i)| = 3\sqrt2.$