If z = a + ib satisfies arg ( z - 1 ) = arg ( z + 3i ) , then ( a -...

We have,

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

$⇒ arg ( ( a - 1 ) + ib ) = arg ( a + ( b + 3 ) i )$

$⇒ \tan^{-1} \left(\dfrac{b}{a - 1}\right) = \tan^{-1}\left(\dfrac{b + 3}{a}\right)</p><p>$⇒ $\dfrac{b}{a - 1} = \dfrac{b + 3}{a}</p><p>$⇒ ab = ( a - 1 ) ( b + 3 )$</p><p>$⇒ 3 ( a - 1 ) = b$</p><p>$⇒ ( a - 1 ) : b = 1: 3$</p>$