Question

Let $z_1=10+6i$ and $z_2=4+6i$. If $z$ is a complex number such that the argument of $(z-z_1)/(z-z_2)$ is $\pi /4$, then prove that $|z-7-9i|=3\sqrt{2}$.

Answer 1

$\frac{\pi }{4}=\arg \frac{z-z_1}{z-z_2}=\arg \left(\frac{x+iy-10-6i}{x+iy-4-6i}\right)$
$\frac{\pi }{4}=arg\frac{z-z_1}{z-z_2}=arg(z-z_1)-arg(z-z_2)$
or $\pi /4={\theta }_1-{\theta }_2$
or ${\tan }^{-1}1={\tan }^{-1}\frac{y_1}{x_1}-{\tan }^{-1}\frac{y_2}{x_2}$
or ${\tan }^{-1}1={\tan }^{-1}\frac{y-6}{x-10}-{\tan }^{-1}\frac{y-6}{x-4}$
or ${\tan }^{-1}1={\tan }^{-1}\frac{(y-6)[\frac{1}{x-10}-\frac{1}{x-4}]}{1+\frac{{(y-6)}^2}{(x-10)(x-4)}}$
$\therefore 1=\frac{6(y-6)}{x^2-14x+40+y^2-12y+36}$
or $x^2-14x+y^2-18y+112=0$
or ${(x-7)}^2-49+{(y-9)}^2-81+112=0$
$\therefore {(x-7)}^2+{(y-9)}^2=18$ ....$\left(1\right)$
$\therefore |z-7-9i|=|(x-7)+i(y-9)|$
$={[{(x-7)}^2+{(y-9)}^2]}^{1/2}=\surd \left(18\right)=3\sqrt{2}$ by $\left(1\right)$