Q11 of 66 Page 91

Solve that equation |z| = z + 1 + 2i.

Given |z| = z + 1 + 2i

Putting z = x + iy, we get

⇒ |x + iy| = x + iy + 1 + 2i

We know that

Comparing real and imaginary parts,

And 0 = y + 2

⇒ y = -2

Putting this value of y in ,

⇒ x² + (-2)² = (x + 1)²

⇒ x² + 4 = x² + 2x + 1

∴ x = 3/2

∴ z = x + iy

= 3/2 – 2i

If the real part of is 4, then show that the locus of the point representing z in the complex plane is a circle.

Show that the complex number z, satisfying the condition arg lies on a circle.

If |z + 1| = z + 2 (1 + i), then find z.

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