If z = x + iy, then area of the triangle whose vertices are points z, iz, z + iz is (1) (1/2)|z|^2 (2) 1/4|z|^2 (3) |z|^2 (4) 3/2 |z|2

Solution:

Let A = z, B = iz, C = z+iz

|AB| = |iz-z|

|BC| = z

|AC| = iz

|AC| = |BC| 

P is the centre of AB

P = (z+iz)/2

Area of triangle ABC = ½ |AB×PC|

= ½ ((z+iz)/2)×(iz-z)

= ¼(i|z|²-|z|²-|z|²-i|z|²)

= |z|²/2

Hence option (1) is the answer.