Question

Find all complex numbers $z$ for which
$\arg \left(\frac{3z-6-3i}{2z-8-6i}\right)=\frac{\pi }{4}$ and $|z-3+4i|=3.$

Answer 1

$\arg \frac{z_1}{z_2}=\arg z_1-{\arg z}_2$
Let $z=x+iy$
$\arg \frac{3(x-2)+i3(y-1)}{2(x-4)+i2(y-3)}=\frac{\pi }{4}$
$\therefore {\tan }^{-1}\frac{y-1}{x-2}-\frac{y-3}{x-4}=\frac{\pi }{4}$
$\therefore \frac{\frac{y-1}{x-2}-\frac{y-3}{x-4}}{1+\frac{(y-1)(y-3)}{(x-2)x-4\left(\right)}}=\tan \frac{\pi }{4}=1$
$2x-2y-2=x^2+y^2-6x-4y+11$
or $x^2+y^2-8x-2y+13=0$ ...$\left(1\right)$
and $|z-3+i|=3$
$\Rightarrow {(x-3)}^2+{\left(y+1\right)}^2=9$
$x^2+y^2-6x+2y+1=0$ ...$\left(2\right)$
Subtracting $\left(1\right)$ and $\left(2\right)$, we get
$x+2y=6\therefore x=6-2y$
Putting in $\left(1\right)$, ${5y}^2-10y+1=0$
$\therefore y=1+\frac{2}{\surd 5},1-\frac{2}{\surd 5}$
$\therefore x=4+\frac{4}{\surd 5},4-\frac{4}{\surd 5}$
$\therefore z=x+iy$ where $(x,y)$ is $(4-\frac{4}{\surd 5},1+\frac{2}{\surd 5})$and $(4+\frac{4}{\surd 5},1-\frac{2}{\surd 5})$