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

If Real ((2z-1)/(z+1)) =1, then locus of z is , where z=x+iy and i=sqrt(-1)

If Real ((2z-1)/(z+1)) =1, then locus of z is , where z=x+iy and i=sqrt(-1)

If Real ((2z-1)/(z+1)) =1, then locus of z is , where z=x+iy and i=sqrt(-1)

Prove that z=i^(i), where i=sqrt(-1), is purely real.

The complex numbers z_(1),z_(2),z_(3) stisfying (z_(2)-z_(3))=(1+i)(z_(1)-z_(3)).where i=sqrt(-1), are vertices of a triangle which is

The complex numbers z_(1),z_(2),z_(3) stisfying (z_(2)-z_(3))=(1+i)(z_(1)-z_(3)).where i=sqrt(-1), are vertices of a triangle which is

If z=(i) ^((i) ^(i)) where i= sqrt (−1) ​ , then z is equal to

Prove that z=i^i, where i=sqrt-1, is purely real.

Prove that z=i^i, where i=sqrt-1, is purely real.