If (z-i/z+i)(z ≠-i) is a purely imaginary number, then z barz is equal to

Question

If (z-i/z+i)(z ≠-i) is a purely imaginary number, then z barz is equal to (A) 0 (B) 1 (C) 2 (D) -1

Tardigrade Admin · Accepted Answer

(z-i/z+i)=λ i (z-i/z+i)+( barz+i/ barz-i)=0 ((z-i)( barz-i)+( barz+i)(z+i)/(z+i)( barz-i))=0 z barz-i barz-i z-1+z barz+i z+i barz-1=0 2 z barz-2=0 z barz=1

Q. If $\frac{z - i}{z + i} (z \neq = - i)$ is a purely imaginary number, then $z \overset{z}{ˉ}$ is equal to

0

1

2

$- 1$

Q. If z+iz−i​(z=−i) is a purely imaginary number, then zzˉ is equal to

0

1

2

−1

z+iz−i​=λi z+iz−i​+zˉ−izˉ+i​=0 (z+i)(zˉ−i)(z−i)(zˉ−i)+(zˉ+i)(z+i)​=0 zzˉ−izˉ−iz−1+zzˉ+iz+izˉ−1=0 2zzˉ−2=0 zzˉ=1

Q. If $\frac{z - i}{z + i} (z \neq = - i)$ is a purely imaginary number, then $z \overset{z}{ˉ}$ is equal to

$- 1$