Q. Number of unimodular complex number which satisfies the locus is

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Solution:


i.e., line segment joining ' ' and '' subtends right angle at variable point
Locus of point is as shown in the figure.
Now, unimodoular complex numbers lie on the circle with center at origin and radius .
image
Clearly, two point ' ' and ' ' are possible points, but these points are not satisfying .
Hence, no such complex number.