We have $\arg (z-1)=\arg (z+3 i)$ and $z=x+i y$
$\Rightarrow \quad \arg (x+i y-1)=\arg (x+i y+3 i)$
$\Rightarrow \quad \arg (x-1+i y)=\arg [x+i(y+3)]$
$\Rightarrow \quad \tan ^{-1} \frac{y}{x-1} \quad=\tan ^{-1} \frac{y+3}{x}$
$$
\begin{aligned}
&\quad\left[\because \arg (z)=\theta=\tan ^{-1} \frac{y}{x}\right] \\
&\Rightarrow \frac{y}{x-1}=\frac{y+3}{x} \\
&\Rightarrow x y=(x-1)(y+3) \\
&\Rightarrow x y=x y-y+3 x-3 \\
&\Rightarrow 3 x-3=y \\
&\Rightarrow \frac{3(x-1)}{y}=1
\end{aligned}
$$
$$
\Rightarrow \frac{x-1}{y}=\frac{1}{3}
$$
Therefore, $(x-1): y=1: 3$