Q13 of 66 Page 91

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

Let z = x + iy


Given arg (z – 1) = arg (z + 3i)


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


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




⇒ xy = xy – y + 3x – 3


⇒ 3x – 3 = y



∴ (x – 1): y = 1: 3


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