Question

if $z=x+iy$ satisfies amp $\left(z-1\right)=amp\left(z+3i\right)$ then the value of $\left(x-1\right):y$ is equal to

Answer 1

$z-1=x+iy-1=(x-1)+iy$ and $z+3i=x+iy+3i=x+i(y+3)$

$arg(z-1)=arg(z+3i)$

$\Rightarrow \arg [(x-1)+iy]=arg[x+i(y+3)]$

$\Rightarrow {\tan }^{-1}\left(\frac{y}{x-1}\right)={\tan }^{-1}\left(\frac{y+3}{x}\right)$

$\Rightarrow {\tan }^{-1}\left(\frac{y}{x-1}\right)-{\tan }^{-1}\left(\frac{y+3}{x}\right)=0$

$\Rightarrow {\tan }^{-1}\left[\frac{\frac{y}{x-1}-\frac{y+3}{x}}{1+\frac{y}{x-1}\frac{y+3}{x}}\right]=0$

$\Rightarrow \frac{xy-(y+3)(x-1)}{(x-1)x+y(y+3)}=0$

$\Rightarrow xy-xy+y-3x+3=0$

$\Rightarrow y=3(x-1)$

$\Rightarrow \frac{x-1}{y}=\frac{1}{3}$

$\therefore x-1:y=1:3$