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B
an ellipse
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C
a parabola
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D
a straight line
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Solution
The correct option is D a straight line Let z=x+iy. Then |z2−1|=|z|2+1⇒|(x2−y2−1)+2ixy|=x2+y2+1⇒(x2−y2−1)2+4x2y2=(x2+y2+1)2⇒(x2−y2−1)2−(x2+y2+1)2+4x2y2=0⇒(2x2)(−2y2−2)+4x2y2=0⇒−4x2=0∴x=0