Question

Let \[u = \dfrac{{2z + i}}{{z - ki}}\], \[z = x + iy\] and \[k > 0\]. If the curve represented by \[Re\left( u \right) + Im\left( u \right) = 1\] intersect the y-axis at the points \[P\] and \[Q\], where \[PQ = 5\]. Then find the value of \[k\]
A. 4
B. \[\dfrac{1}{2}\]
C. 2
D. \[\dfrac{3}{2}\]

Answer 1

Hint: In the given question, two complex equations are given. We will put \[z = x + iy\] in the equation \[u = \dfrac{{2z + i}}{{z - ki}}\] and simplify it in the form of \[a + ib\]. Then add the real part and imaginary part and equate to \[1\]. We will put \[x = 0\]in the equation \[Re\left( u \right) + Im\left( u \right) = 1\] to find the \[y\] intercept. Then apply the formula sum of roots and product of roots. Assume the coordinate of \[P\] and \[Q\]. We find the distance between two points by using the formula \[d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} \] and equate with 5 and square both sides of it. Then by using \[{y_1}^2 + {y_2}^2 = 1 - 2{y_1}{y_2}\] and \[{y_1}{y_2} = - k - {k^2}\], we calculate the value of \[k\]
Formula used :
The conjugate of a complex number \[z = a + ib\] is \[\overline z = a - ib\].
The product of a complex number and its conjugate is: \[\left( {a + ib} \right)\left( {a - ib} \right) = \left( {{a^2} + {b^2}} \right)\]

Distance formula:
The distance between the two points \[\left( {{x_1},{y_1}} \right)\] and \[\left( {{x_2},{y_2}} \right)\] is: \[d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} \]

If \[\alpha \] and \[\beta \] are the roots of a quadratic equation \[a{x^2} + bx + c = 0, a \ne 0\], then
\[\alpha + \beta = - \dfrac{b}{a}\] and \[\alpha \beta = \dfrac{c}{a}\]
Complete step by step solution:
The given complex equations are \[u = \dfrac{{2z + i}}{{z - ki}}\] and \[z = x + iy\].
Let’s simplify the complex equation \[u = \dfrac{{2z + i}}{{z - ki}}\].
Substitute \[z = x + iy\] in above equation.
\[u = \dfrac{{2\left( {x + iy} \right) + i}}{{\left( {x + iy} \right) - ki}}\]
\[ \Rightarrow \]\[u = \dfrac{{2x + i\left( {2y + 1} \right)}}{{x + i\left( {y - k} \right)}}\]
Now multiply the numerator and denominator of the above equation by the complex conjugate of the denominator.
\[u = \dfrac{{2x + i\left( {2y + 1} \right)}}{{x + i\left( {y - k} \right)}} \times \dfrac{{x - i\left( {y - k} \right)}}{{x - i\left( {y - k} \right)}}\]
Simplify the above equation.
\[u = \dfrac{{2{x^2} - {i^2}\left( {2y + 1} \right)\left( {y - k} \right) + i\left( {2xy + x} \right) - i\left( {2xy - 2xk} \right)}}{{{x^2} + {{\left( {y - k} \right)}^2}}}\]
\[ \Rightarrow \]\[u = \dfrac{{2{x^2} + \left( {2y + 1} \right)\left( {y - k} \right) + i\left( {2xy + x - 2xy + 2xk} \right)}}{{{x^2} + {{\left( {y - k} \right)}^2}}}\] [ Since \[{i^2} = - 1\]]
\[ \Rightarrow \]\[u = \dfrac{{2{x^2} + \left( {2y + 1} \right)\left( {y - k} \right) + i\left( {x + 2xk} \right)}}{{{x^2} + {{\left( {y - k} \right)}^2}}}\]
\[ \Rightarrow u = \dfrac{{2{x^2} + \left( {2y + 1} \right)\left( {y - k} \right)}}{{{x^2} + {{\left( {y - k} \right)}^2}}} - i\dfrac{{\left( {x + 2xk} \right)}}{{{x^2} + {{\left( {y - k} \right)}^2}}}\]
The given equation of a curve intersecting the y-axis at the points \[P\] and \[Q\] is,
\[Re\left( u \right) + Im\left( u \right) = 1\]
\[ \Rightarrow \]\[\dfrac{{2{x^2} + \left( {2y + 1} \right)\left( {y - k} \right)}}{{{x^2} + {{\left( {y - k} \right)}^2}}} + \dfrac{{\left( {x + 2xk} \right)}}{{{x^2} + {{\left( {y - k} \right)}^2}}} = 1\]
\[ \Rightarrow \]\[2{x^2} + \left( {2y + 1} \right)\left( {y - k} \right) + \left( {x + 2xk} \right) = {x^2} + {\left( {y - k} \right)^2}\]
Since the curve intersect the y-axis.
So, at y-axis, \[x = 0\].
Substitute \[x = 0\] in the above equation.
\[2{\left( 0 \right)^2} + \left( {2y + 1} \right)\left( {y - k} \right) + \left( {0 + 2\left( 0 \right)k} \right) = {\left( 0 \right)^2} + {\left( {y - k} \right)^2}\]
\[ \Rightarrow \]\[\left( {2y + 1} \right)\left( {y - k} \right) = {\left( {y - k} \right)^2}\]
\[ \Rightarrow \]\[2{y^2} + y - 2yk - k = {y^2} - 2yk + {k^2}\]
\[ \Rightarrow \]\[{y^2} + y - \left( {k + {k^2}} \right) = 0\]
The roots of the above equation are the y-coordinates of the points \[P\] and \[Q\].
Let \[{y_1}\] and \[{y_2}\] are the roots of the above equation. Then
\[{y_1} + {y_2} = - 1\] and \[{y_1}{y_2} = - k - {k^2}\]
Take square of the first equation.
\[{y_1}^2 + {y_2}^2 + 2{y_1}{y_2} = 1\]

The distance between the points \[P\left( {0,{y_1}} \right)\] and \[Q\left( {0,{y_2}} \right)\] is 5.
Apply the distance formula \[d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} \].
\[5 = \sqrt {{{\left( {0 - 0} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} \]
Simplify the above equation.
\[5 = \left| {{y_2} - {y_1}} \right|\]
Take square on both sides.
\[25 = {y_2}^2 - 2{y_1}{y_2} + {y_1}^2\]
\[ \Rightarrow \]\[25 = 1 - 2{y_1}{y_2} - 2{y_1}{y_2}\] [ Since \[{y_1}^2 + {y_2}^2 = 1 - 2{y_1}{y_2}\] ]
\[ \Rightarrow \]\[25 = 1 - 4{y_1}{y_2}\]
\[ \Rightarrow \]\[25 = 1 + 4\left( {{k^2} + k} \right)\] [ Since \[{y_1}{y_2} = - k - {k^2}\]]
\[ \Rightarrow \]\[24 = 4\left( {{k^2} + k} \right)\]
\[ \Rightarrow \]\[6 = {k^2} + k\]
\[ \Rightarrow \]\[{k^2} + k - 6 = 0\]
Factorize the above equation.
\[{k^2} + 3k - 2k - 6 = 0\]
\[ \Rightarrow \]\[k\left( {k + 3} \right) - 2\left( {k + 3} \right) = 0\]
\[ \Rightarrow \]\[\left( {k + 3} \right)\left( {k - 2} \right) = 0\]
\[ \Rightarrow \]\[k + 3 = 0\] or \[k - 2 = 0\]
\[ \Rightarrow \]\[k = - 3\] or \[k = 2\]

It is given that \[k > 0\].
So, the correct solution is \[k = 2\].
Hence the correct option is C..
Note: Students are often skip the point that \[P\] and \[Q\] lie on the \[y\]axis. So the coordinates of the points should be in the form \[\left( {0,a} \right)\]. They assume the coordinate in the form \[\left( {a,b} \right)\] and stuck in that place.