# If z2

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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