BITSAT 2019 Paper

(a) Let z = x + iy
∴ | z + 3 - i | = |(x + 3) + i(y - 1)| = 1
⇒ √(x+3)2+(y−1)2 = 1 ...(i)
∵ arg z = π ⇒ tan−1

y

x

= π

y

x

= tan π = 0 ⇒ y = 0 ...(ii)
From equations (i) and (ii), we get
x = -3, y = 0 ∴ z = -3
⇒ |z| = |-3| = 3