If `z and w` are two complex number such that `|zw|=1 and arg (z) – arg (w) = pi/2,` then show that `overline zw = -i.`

Options

A. `bar z w =i`
B. `zbarw=(1-i)/sqrt(2)`
C. `bar z w=i`
D. `z bar w = (-1 +i)/sqrt(2)`

Answer

A

Explanation

It is given that there, are two complex numbers z and w sunc that `|z w|=1 and arg(w) = pi//2`
`therefore |z||w|=1 [ because |z_1 z_2|= | z_1||z_2|]`
and `arg(z) = pi/2 + arg (w)`
Let `|z| =r , then | w|= 1/r`
and let arg (w) `= theta " then arg(z) pi/2 + theta `
So we can assume
`z= re^(i pi//2+ theta)`
`[ because ` if z= x+ iy is a complex number then it can be written as z `=re^(i theta) where r= |z| and theta = arg (z) ]`
and `w =1/re^(it theta)`
Now ` barz.w = re^(-i(pi//2theta)). 1/re^(i theta)`
`=e^(i(-pi//2- theta+ theta))=e^(-i(pi//2))-i " " [ because e^(-i theta) = cos theta - i sin theta]`
and `z bar w = re^(i(pi//2 + theta)).1/r e^(- i theta)`
`= e^(i(pi//2+ theta - theta))=e^(i(pi//2))=i`