Let u=2z+iz-ki, z=x+iy and k0.

Question

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

Accepted Answer

Given, z = x + iy

and $$u = {{2z + i} \over {z - ki}}$$

$$ = {{2(x + iy) + i} \over {(x + iy) - ki}}$$

$$ = {{2x + i(2y + 1)} \over {x + i(y - k)}} \times {{x - i(y - k)} \over {x - i(y - k)}}$$

$$ = {{2{x^2} + (2y + 1)(y - k) + i(2xy + x - 2xy + 2kx)} \over {{x^2} + {{(y - k)}^2}}}$$

Given,

$${\mathop{\rm Re}\nolimits} (u) + {\mathop{\rm Im}\nolimits} (u) = 1$$

$$ \Rightarrow {{2{x^2} + (2y + 1)(y - k)} \over {{x^2} + {{(y - k)}^2}}} + {{x + 2kx} \over {{x^2} + {{(y - k)}^2}}} = 1$$

$$ \Rightarrow 2{x^2} + (2y + 1)(y - k) + x + 2kx = {x^2} + {(y - k)^2}$$

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,

$$ \therefore $$ $(2y + 1)(y - k) = {(y - k)^2}$

$$ \Rightarrow 2{y^2} + y - 2yk - k = {y^2} + {k^2} - 2ky$$

$$ \Rightarrow {y^2} + y - (k + {k^2}) = 0$$

Let roots of this quadratic equation y1 and y2

$$ \therefore $$ Point P (0, y1) and Q (0, y2)

and ${y_{_1}} + {y_2} = 1$ , ${y_1}{y_2} =  - k - {k^2}$

$$ \therefore $$ $${({y_1} - {y_2})^2} = {({y_1} + {y_2})^2} - 4{y_1}{y_2}$$

$= 1 + 4k + 4{k^2}$

$$ \Rightarrow |{y_1} - {y_2}| = \sqrt {1 + 4k + 4{k^2}} $$

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)

Solutions · 1