Cierra Castillo

2022-07-12

If argument of $\frac{z-{z}_{1}}{z-{z}_{2}}$ is $\frac{\pi }{4}$, 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?

Jenna Farmer

Expert

the angle subtended by the chord ${z}_{1}{z}_{2}$ 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 $\frac{-6+4i}{2}$ so that you will get two centers. the two centres, ${z}_{1}$ and ${z}_{2}$ form a square of side $\sqrt{26}.$

Savanah Boone

Expert

Put $\phantom{\rule{thickmathspace}{0ex}}z=x+iy\phantom{\rule{thickmathspace}{0ex}},\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}x,y\in \mathbb{R}\phantom{\rule{thickmathspace}{0ex}}$, so
$\frac{z-2-3i}{z-6-9i}=\frac{\left(x-2\right)+\left(y-3\right)i}{\left(x-6\right)+\left(y-9\right)i}\cdot \frac{\left(x-6\right)-\left(y-9\right)i}{\left(x-6\right)-\left(y-9\right)i}=$
$=\frac{\left(x-2\right)\left(x-6\right)+\left(y-3\right)\left(y-9\right)}{\left(x-6{\right)}^{2}+\left(y-9{\right)}^{2}}+\frac{\left(x-6\right)\left(y-3\right)-\left(x-2\right)\left(y-9\right)}{\left(x-6{\right)}^{2}+\left(y-9{\right)}^{2}}i$
By the given data, it must be that the real and imaginary parts are identical, and thus
$\left(x-2\right)\left(x-6\right)+\left(y-3\right)\left(y-9\right)=\left(x-6\right)\left(y-3\right)-\left(x-2\right)\left(y-9\right)\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}$
$\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}{x}^{2}-14x+{y}^{2}-8y-26=0$
Complete squares, make some algebraic hokus pokus and get a circle.

Do you have a similar question?