` sin^(-1) {1/i ( z-1)}`, where z is non real and `i = sqrt(-1)`, can be the angle of a triangle If:

Options

A. Re(z)=1, Im(z)=2
B. Re(z)=1,`-1 le "Im"(z) le 1`
C. Re(z)+Im(z)=0
D. None of these

Answer

B

Explanation

For `sin^(-1)(z-1)/i` to be the angle of a triangle `(z-1)/i` must be real and `-1 le(z-1)/I le 1`
This will be real, if `x-1=0` i.e., `x=1`
Also, `-1 le(z-1)/I le 1 rArr -1 le y le 1`
Thus, Re(z)=1 and `-1 le "Im"(z) le 1`.