Question

Find the complex numbers $z$ which simultaneously satisfy the equations

$\left|\frac{z-12}{z-8i}\right|=\frac{5}{3}$ and $\left|\frac{z-4}{z-8}\right|=1$.

Answer 1

Putting $z=x+iy$, the given equation
become $\left|\frac{x+iy-12}{x+iy-8i}\right|=\frac{5}{3}$ and $\left|\frac{x+iy-4}{x+iy-8}\right|=1$
or $9{|(x-12)+iy|}^2=25{|x+i(y-8)|}^2$
and ${|(x-4)+iy|}^2={|(x-8)+iy|}^2$
or $9[{(x-12)}^2+y^2]=25[x^2+{(y-8)}^2]$
and ${(x-4)}^2+y^2={(x-8)}^2+y^2$
The above equation can be written as
${\left[3(x-12)\right]}^2-{\left(5x\right)}^2={\left[5(y-8)\right]}^2-{\left(3y\right)}^2$
and ${(x-4)}^2-{(x-8)}^2=0$
Apply $a^2-b^2=(a+b)(a-b)$
$(8x-36)(-2x-36)=(8y-40)(2y-40)$ .....$\left(1\right)$
and $(2x-12)\left(4\right)=0$ $\therefore x=6$ .....$\left(2\right)$
and putting in (1), we get
$12(-48)=8\times 2(y-5)(y-20)$
or $y^2-25y+136=0$ or $(y-8)(y-17)=0$
$\therefore y=8,17$
Hence $z_1=(6,8)$ and $z_2=(6,17)$
i.e. $z_1=6+8i$ and $z_2=6+17i$ are two solutions.