JEE Main 2020 · 4 September · morning · Single correct · +4 / −1

Let u = (2z + i)/(z - ki), z = x + iy and k > 0. If the curve represented
by Re(u) + Im(u) = 1 intersects the y-axis at the points P and Q where PQ = 5, then the value of k is :

  1. (a) 2
  2. (b) 4
  3. (c) 1/2
  4. (d) 3/2
Answer and solution

Answer (a) 2

Solution

Maths sir · 4 steps

Clear the denominator to get a circle, then put x = 0 and read off the two y-intercepts.

  1. Split u. Write u = (2x + i(2y+1))/(x + i(y-k)) and multiply top and bottom by x - i(y-k). This gives Re(u) = (2x² + (2y+1)(y-k))/D and Im(u) = (1+2k)x/D, where D = x² + (y-k)².
  2. Form the curve. Set Re(u) + Im(u) = 1 and multiply through by D: 2x² + (2y+1)(y-k) + (1+2k)x = x² + (y-k)². Expanding and cancelling the -2ky terms leaves a circle.

    x² + y² + (1+2k)x + y - k(k+1) = 0

  3. Cut the y-axis. Put x = 0: y² + y - k(k+1) = 0, which factors as (y - k)(y + k + 1) = 0. So P and Q are at y = k and y = -(k+1).
  4. So. PQ = k - (-(k+1)) = 2k + 1 = 5, so k = 2, which is option (a).

The idea to keep. For a locus like Re(u) + Im(u) = 1, put z = x + iy, split u into real and imaginary parts and clear the denominator. A chord on an axis is the difference of the two roots you get there.

Answer from the NTA final answer key. Solution by PrepWiser.

Study Complex Numbers

Asked about 160 times in JEE Main since 2020. This question is from 2020.