∵z2=z⋅21−|z|......‌ (1) ‌ ⇒|z|2=|z|⋅21−|z| ⇒|z|=21−|z| ∵b≠0⇒|z|≠0 ∴|z|=1......‌ (2) ‌ ∵z=a+ib‌ then ‌√a2+b2=1......‌ (3) ‌ Now again from equation (1), equation (2), equation (3) we get : a2−b2+i2ab=(a−ib)20 ∴a2−b2=a‌ and ‌2ab=−b ∴a=−‌