Let be such that . If , then
- A.
- B.
- C.
- D.
Show answer and solution
Correct option: B
Solution
Let z = x + iy with x² + y² < 1. Write ω = (5 + 3z)/(5(1-z)).
Re(ω) = Re[(5 + 3z)/(5(1-z))]. Let z = x+iy, then:
5(1-z) = 5(1-x) - 5iy, and 5+3z = (5+3x)+3iy.
Re(ω) = [(5+3x)·5(1-x) + 3y·5y] / [25(1-x)² + 25y²]
= [5(5+3x)(1-x) + 15y²] / [25((1-x)² + y²)]
= [(5+3x)(1-x) + 3y²] / [5((1-x)² + y²)]
Numerator: (5+3x)(1-x) + 3y² = 5 - 5x + 3x - 3x² + 3y² = 5 - 2x - 3(x²+y²)
Denominator: 5[(1-x)² + y²] = 5[1 - 2x + x² + y²]
So 5·Re(ω) = (5 - 2x - 3(x²+y²))/(1 - 2x + x² + y²).
Let r² = x²+y² < 1. We need to show 5·Re(ω) > 1:
(5 - 2x - 3r²) > (1 -
Re(ω) = Re[(5 + 3z)/(5(1-z))]. Let z = x+iy, then:
5(1-z) = 5(1-x) - 5iy, and 5+3z = (5+3x)+3iy.
Re(ω) = [(5+3x)·5(1-x) + 3y·5y] / [25(1-x)² + 25y²]
= [5(5+3x)(1-x) + 15y²] / [25((1-x)² + y²)]
= [(5+3x)(1-x) + 3y²] / [5((1-x)² + y²)]
Numerator: (5+3x)(1-x) + 3y² = 5 - 5x + 3x - 3x² + 3y² = 5 - 2x - 3(x²+y²)
Denominator: 5[(1-x)² + y²] = 5[1 - 2x + x² + y²]
So 5·Re(ω) = (5 - 2x - 3(x²+y²))/(1 - 2x + x² + y²).
Let r² = x²+y² < 1. We need to show 5·Re(ω) > 1:
(5 - 2x - 3r²) > (1 -
[JEE Main 2019 · 9 April · Shift II]