Q7 of 202 Page 188

If z2 + |z|2 = 0, show that z is purely imaginary.

Let z= a + ib


⇒ |z| = √(a2 + b2)


Now , z2 + |z|2 = 0


⇒ (a + ib)2 + a2 + b2 = 0


⇒ a2 + 2abi + i2b2 + a2 + b2 = 0


⇒ a2 + 2abi - b2 + a2 + b2 = 0


⇒ 2a2 + 2abi = 0


⇒ 2a(a + ib) = 0


Either a = 0 or z = 0


Since z≠ 0


a = 0 ⇒ z is purely imaginary.


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