If arg (z – 1) = arg (z + 3i),

According to the question,

Let z = x + iy

Given that,

arg (z – 1) = arg (z + 3i)

⇒ arg (x + iy – 1) = arg (x + iy + 3i)

⇒ arg (x – 1 + iy) = arg (x + I (y) = π/4

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

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

⇒ xy = xy – y + 3x – 3

⇒ 3x – 3 = y

⇒ (x – 1)/y = 1/3

Hence, (x – 1): y = 1: 3