Let u = {{2z + i} \over {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 :

Question

Let u = {{2z + i} \over {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 :

Accepted Answer

Answer: A. 2 Given, z = x + iy and u = {{2z + i} \over {z - ki}} = {{2(x + iy) + i} \over {(x + iy) - ki}} = {{2x + i(2y + 1)} \over {x + i(y - k)}} 	imes {{x - i(y - k)} \over {x - i(y - k)}} = {{2{x^2} + (2y + 1)(y - k) + i(2xy + x - 2xy + 2kx)} \over {{x^2} + {{(y - k)}^2}}} Given, {\mathop{m Re}
olimits} (u) + {\mathop{m Im}
olimits} (u) = 1 \Rightarrow {{2{x^2} + (2y + 1)(y - k)} \over {{x^2} + {{(y - k)}^2}}} + {{x + 2kx} \over {{x^2} + {{(y - k)}^2}}} = 1 \Rightarrow 2{x^2} + (2y + 1)(y - k) + x + 2kx = {x^2} + {(y - k)^2} This curve intersect the y-axis at point P and Q, so at point P and Q x = 0 Putting x = 0 at the above equation, 	herefore (2y + 1)(y - k) = {(y - k)^2} \Rightarrow 2{y^2} + y - 2yk - k = {y^2} + {k^2} - 2ky \Rightarrow {y^2} + y - (k + {k^2}) = 0 Let roots of this quadratic equation y1 and y2 	herefore Point P (0, y1) and Q (0, y2) and {y_{_1}} + {y_2} = 1 , {y_1}{y_2} = - k - {k^2} 	herefore {({y_1} - {y_2})^2} = {({y_1} + {y_2})^2} - 4{y_1}{y_2} = 1 + 4k + 4{k^2} \Rightarrow |{y_1} - {y_2}| = \sqrt {1 + 4k + 4{k^2}} Given, PQ = 5 \Rightarrow |{y_1} - {y_2}| = 5 \Rightarrow \sqrt {1 + 4k + 4{k^2}} = 5 \Rightarrow {k^2} + k - 6 = 0 \Rightarrow k = - 3,\,2 So, k = 2 (Given k > 0)

Let $$u = {{2z + i} \over {z - ki}}$$, 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 :

Answer

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