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If $arg (z - 1) = arg (z + 3i)$ , then find x – 1 : y. where $z = x + iy$

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

$arg (x+iy-1)=arg(x+iy+3i)$

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

$\tan^{-1}\frac{y}{x-1}=\tan^{-1}\frac{y+3}{x}$

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

$xy=xy-y+3x-3$

$3x-3=y$

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

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