For every real number c≥0, find all the complex numbers z which satisfy the equation 2|z|−4cz+1+ic=0.
A
(x,y)=4c±√4c2+34(4c2−1),1/4
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
(x,y)=4c±√4c2+34(4c2+1),−1/4
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
(x,y)=4c±√4c2+34(4c2−1),−1/4
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
(x,y)=4c±√4c2+34(4c2+1),1/4
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is A(x,y)=4c±√4c2+34(4c2−1),1/4 Let z=x+iy ⇒2√x2+y2−4c(x+iy)+1+ic=0 Considering the imaginary part, we get ⇒−4cy+c=0 ∴y=14 Substituting in the real part, we get ⇒√16x2+1−8cx+2=0