Q.
Number of unimodular complex number which satisfies the locus arg(z+iz−1)=2π is
262
149
Complex Numbers and Quadratic Equations
Report Error
Solution:
arg(z+iz−1)=2π
i.e., line segment joining ' 1 ' and '−i' subtends right angle at variable point P(z)
Locus of point P(z) is C1 as shown in the figure.
Now, unimodoular complex numbers lie on the circle C2 with center at origin and radius 1.
Clearly, two point ' 1 ' and ' −i ' are possible points, but these points are not satisfying (1).
Hence, no such complex number.