The real value of θ for which the expression, 1+icosθ1−2icosθ is real number is
A
nπ±π2
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
nπ−π2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
nπ+π2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
2nπ±π2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is Anπ±π2 Let z=1+icosθ1−2icosθ =(1+icosθ)(1+2icosθ)(1−2icosθ)(1+2icosθ) =1−2cos2θ+i3cosθ1+4cos2θ =1−2cos2θ1+4cos2θ+i(3cosθ1+4cos2θ) If the above complex number is purely real, then Im(z)=0 3cosθ=0 θ=(2n±1)π2, where nϵN