Let ˙z=x+iy z‌=x−iy⇒z=iz2 ⇒x−iy‌=i[(x+iy)(x+iy)] ⇒x−iy‌=i[x2+2ixy+i2y2] ⇒x−iy‌=i[x2−y2+2ixy] ⇒x−iy‌=−2xy+i(x2−y2) Comparing the real and imaginary part, we get −2xy=x ⇒‌ Either ‌x‌=0‌ or ‌y=−‌
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x2−y2‌=−y⇒x2=y2−y If x=0, then y=0 or y=1. If y=−‌