Let be a complex number such that (where ). Then, is equal to
- A.
- B.
- C.
- D.
Show answer and solution
Correct option: B
Solution
Let z = x + iy. Then |z| = √(x²+y²) is real, so the imaginary part of |z|+z equals y = 1, giving y = 1.
The real part gives √(x²+y²) + x = 3, so √(x²+1) = 3-x. Squaring: x²+1 = 9-6x+x², giving 6x = 8, x = 4/3.
Therefore |z| = √((4/3)²+1²) = √(16/9 + 9/9) = √(25/9) = 5/3.
The real part gives √(x²+y²) + x = 3, so √(x²+1) = 3-x. Squaring: x²+1 = 9-6x+x², giving 6x = 8, x = 4/3.
Therefore |z| = √((4/3)²+1²) = √(16/9 + 9/9) = √(25/9) = 5/3.
[JEE Main 2019 · 11 Jan · Shift II]