Question

The set of all $\alpha \in R$, for which $\omega = \dfrac{{1 + \left( {1 - 8\alpha } \right)z}}{{1 - z}}$ is a purely imaginary number, for all $z \in C$ satisfying $\left| z \right| = 1$ and $\operatorname{Re} z \ne 1$, is
\left( A \right)\left\\{ 0 \right\\}
$\left( B \right)$ An empty set
\left( C \right)\left\\{ {0,\dfrac{1}{4},\dfrac{{ - 1}}{4}} \right\\}
$\left( D \right)$ Equal to R

Answer 1

Hint – In this particular type of question first assume any complex number (say z = x + iy), so the modulus of z is given as, $\left| z \right| = \left| {x + iy} \right| = \sqrt {{x^2} + {y^2}}$, then use the concept of rationalization, i.e. we have to multiply and divide by conjugate of the denominator so use these concepts to reach the solution of the question.

Complete step by step solution:
Given data:
$\omega = \dfrac{{1 + \left( {1 - 8\alpha } \right)z}}{{1 - z}}$
$z \in C$, where C is a complex number
$\left| z \right| = 1$
$\operatorname{Re} z \ne 1$
Now we have to find the set of all $\alpha \in R$ such that $\omega$ is a purely imaginary number.
First of all let, z = x + iy....................... (1)
So it is given that, $\left| z \right| = 1$
Therefore, $\left| z \right| = \left| {x + iy} \right| = \sqrt {{x^2} + {y^2}} = 1$
Now take square on both sides we have,
$\Rightarrow {\left( {\sqrt {{x^2} + {y^2}} } \right)^2} = {1^2}$
$\Rightarrow {x^2} + {y^2} = 1$.................. (2)
Now it is also given that, $\operatorname{Re} z \ne 1$
So from equation (1), the real part of z is x and the imaginary part of z is y.
Therefore, $x \ne 1$........ (3)
Now given equation is, $\omega = \dfrac{{1 + \left( {1 - 8\alpha } \right)z}}{{1 - z}}$
Now substitute the value of z in the above equation we have,
$\Rightarrow \omega = \dfrac{{1 + \left( {1 - 8\alpha } \right)\left( {x + iy} \right)}}{{1 - \left( {x + iy} \right)}}$
Now separate real and imaginary part of the denominator of the above equation we have,
$\Rightarrow \omega = \dfrac{{1 + \left( {1 - 8\alpha } \right)\left( {x + iy} \right)}}{{\left( {1 - x} \right) - iy}}$
Now rationalize the above equation we have, (i.e. multiply and divide by the conjugate of the denominator term)
$\Rightarrow \omega = \dfrac{{1 + \left( {1 - 8\alpha } \right)\left( {x + iy} \right)}}{{\left( {1 - x} \right) - iy}} \times \dfrac{{\left( {1 - x} \right) + iy}}{{\left( {1 - x} \right) + iy}}$
Now simplify the above equation we have,
$\Rightarrow \omega = \dfrac{{\left( {1 + x + iy - 8\alpha x - 8i\alpha y} \right)}}{{\left( {1 - x} \right) - iy}} \times \dfrac{{\left( {1 - x} \right) + iy}}{{\left( {1 - x} \right) + iy}}$
$\Rightarrow \omega = \dfrac{{\left[ {\left( {1 + x - 8\alpha x} \right) + i\left( {y - 8\alpha y} \right)} \right]\left[ {\left( {1 - x} \right) + iy} \right]}}{{\left[ {\left( {1 - x} \right) - iy} \right]\left[ {\left( {1 - x} \right) + iy} \right]}}$
Now again simplify the above equation we have,
$\Rightarrow \omega = \dfrac{{\left( {1 + x - 8\alpha x} \right)\left( {1 - x} \right) + {i^2}y\left( {y - 8\alpha y} \right) + iy\left( {1 + x - 8\alpha x} \right) + i\left( {y - 8\alpha y} \right)\left( {1 - x} \right)}}{{{{\left( {1 - x} \right)}^2} - {i^2}{y^2}}}$
Now as we know in complex the value of $i = \sqrt { - 1} \Rightarrow {i^2} = - 1$ so substitute this value in the above equation we have,
$\Rightarrow \omega = \dfrac{{\left( {1 + x - 8\alpha x} \right)\left( {1 - x} \right) + \left( { - 1} \right)y\left( {y - 8\alpha y} \right) + iy\left( {1 + x - 8\alpha x} \right) + i\left( {y - 8\alpha y} \right)\left( {1 - x} \right)}}{{{{\left( {1 - x} \right)}^2} - \left( { - 1} \right){y^2}}}$
Now again simplify and separate real and imaginary terms we have,
$\Rightarrow \omega = \dfrac{{\left( {1 + x - 8\alpha x} \right)\left( {1 - x} \right) - y\left( {y - 8\alpha y} \right) + iy\left( {1 + x - 8\alpha x} \right) + i\left( {y - 8\alpha y} \right)\left( {1 - x} \right)}}{{{{\left( {1 - x} \right)}^2} + {y^2}}}$

$\Rightarrow \omega = \dfrac{{\left( {1 + x - 8\alpha x} \right)\left( {1 - x} \right) - y\left( {y - 8\alpha y} \right)}}{{{{\left( {1 - x} \right)}^2} + {y^2}}} + i\dfrac{{\left( {y - 8\alpha y} \right)\left( {1 - x} \right) + y\left( {1 + x - 8\alpha x} \right)}}{{{{\left( {1 - x} \right)}^2} + {y^2}}}$
Now we have to find the set of all $\alpha \in R$ such that $\omega$ is a purely imaginary number.
Therefore, the real part of the above equation should be zero.
$\operatorname{Re} \left( \omega \right) = \dfrac{{\left( {1 + x - 8\alpha x} \right)\left( {1 - x} \right) - y\left( {y - 8\alpha y} \right)}}{{{{\left( {1 - x} \right)}^2} + {y^2}}} = 0$
Now simplify this this equation we have,
$\Rightarrow \left( {1 + x - 8\alpha x} \right)\left( {1 - x} \right) - y\left( {y - 8\alpha y} \right) = 0$
$\Rightarrow 1 - x + x - {x^2} - 8\alpha x + 8\alpha {x^2} - {y^2} + 8\alpha {y^2} = 0$
$\Rightarrow \left( {1 - x} \right) + x\left( {1 - 8\alpha } \right) - {x^2}\left( {1 - 8\alpha } \right) - {y^2}\left( {1 - 8\alpha } \right) = 0$
$\Rightarrow \left( {1 - x} \right) + x\left( {1 - 8\alpha } \right) - \left( {1 - 8\alpha } \right)\left[ {{x^2} + {y^2}} \right] = 0$
Now from equation (2), ${x^2} + {y^2} = 1$ so we have,
$\Rightarrow \left( {1 - x} \right) + x\left( {1 - 8\alpha } \right) - \left( {1 - 8\alpha } \right) = 0$
$\Rightarrow \left( {1 - x} \right) - \left( {1 - 8\alpha } \right)\left[ {1 - x} \right] = 0$
$\Rightarrow \left( {1 - x} \right)\left[ {1 - \left( {1 - 8\alpha } \right)} \right] = 0$
Now from equation (3), $x \ne 1$
Therefore, $1 - \left( {1 - 8\alpha } \right) = 0$
$\Rightarrow 1 - 1 + 8\alpha = 0$
$\Rightarrow 8\alpha = 0$
$\Rightarrow \alpha = 0$
Therefore, \alpha \in \left\\{ 0 \right\\}
Hence option (A) is the correct answer.

Note – Whenever we face such types of questions the key concept we have to remember is that always recall that if a complex number is purely imaginary then the real part of that complex number should be zero, so first separate the real and imaginary terms of given equation and then equate the real term of this equation to zero and simplify as above we will get the required answer.