Complex number - locus of a point

Question

Question:

If argument of $\frac{z - z_1}{z-z_2}$ is $\pi\over4$, find the locus of $z$. $$z_1 = 2 + 3i$$$$z_2 = 6 + 9i$$

Approach: I tried to solve the equation using diagram, basically plotting the points on the Argand plane. What I got is a circle with center $7 + 4i$ and a radius of $\sqrt{26}$ units. The two complex numbers given lie on this circle, and form a chord. Any point lying on the major arc of this chord satisfies the condition.

How exactly would I represent this as a locus of the point? And is there any other method that I can use that does not involve a diagram?

Answer 1

Put $\;z=x+iy\;,\;\;x,y\in\Bbb R\;$ , so

$$\frac{z-2-3i}{z-6-9i}=\frac{(x-2)+(y-3)i}{(x-6)+(y-9)i}\cdot\frac{(x-6)-(y-9)i}{(x-6)-(y-9)i}=$$

$$=\frac{(x-2)(x-6)+(y-3)(y-9)}{(x-6)^2+(y-9)^2}+\frac{(x-6)(y-3)-(x-2)(y-9)}{(x-6)^2+(y-9)^2}i$$

By the given data, it must be that the real and imaginary parts are identical, and thus

$$(x-2)(x-6)+(y-3)(y-9)=(x-6)(y-3)-(x-2)(y-9)\iff $$

$$\iff x^2-14x+y^2-8y-26=0$$

Complete squares, make some algebraic hokus pokus and get a circle.

Answer 2

the angle subtended by the chord $z_1z_2$ at the center is $2 \pi/4 = \pi/2$ so the radius is $\frac{|z_1-z_2|}{\sqrt 2} = \sqrt{26}$ the center of the chord is $4 + 3i$ you add or subtract $\dfrac{-6+4i}{2}$ so that you will get two centers. the two centres, $z_1$ and $z_2$ form a square of side $\sqrt{ 26}.$