Let z≠-i be any complex number such that (z-i/z+i) is a purely imaginary number. Then z+(1/z) is :

Question

Let z≠-i be any complex number such that (z-i/z+i) is a purely imaginary number. Then z+(1/z) is : (A) 0 (B) any non-zero real number other than 1 (C) any non-zero real number (D) a purely imaginary number

Tardigrade Admin · Accepted Answer

Let z=x+iy (z-i/z+i) is purely imaginary means its real part is zero. (x+iy-i/x+iy+i)=(x + i(y -1)/x + i(y +1))×(x - i(y +1)/x - i(y +1)) =(x2-2ix(y+1)+xi(y-1)+y2-1/x2+(y+1)2) =(x2+y2-1/x2+(y+1)2)-(2xi/x2+(y+)2) for pure imaginary, we have (x2+y2-1/x2+(y+1)2)=0 ⇒ x2+y2=1 ⇒ (x + iy) (x - iy) = 1 ⇒ x+iy=(1/x-iy)=z and (1/z)=x-iy z+(1/z)=(x+iy)+(x-iy)=2x (z+(1/z)) is any non-zero real number

Q. Let z=−i be any complex number such that z+iz−i​ is a purely imaginary number. Then z+z1​ is :

0

any non-zero real number other than 1

any non-zero real number

a purely imaginary number

Q. Let $z \neq = - i$ be any complex number such that $\frac{z - i}{z + i}$ is a purely imaginary number. Then $z + \frac{1}{z}$ is :