Question

Let z be a complex number such that the imaginary part of z is nonzero and $ a = {z^2} + z + 1 $ is real. Then a cannot take the value:
 $
  A) - 1 \\
  B)\dfrac{1}{3} \\
  C)\dfrac{1}{2} \\
  D)\dfrac{3}{4} \\
  $

Answer 1

Hint: In this question, first of all we are going to put the value of z in the equation $ a = {z^2} + z + 1 $ and then simplify the equation. After simplifying, we are going to take an imaginary part of that simplified equation as 0 and find the value of x. After finding the value we can easily get the value of a.

Complete step by step solution:
Given data:
z is a complex number and the imaginary part in not zero.
 $ \Rightarrow z = x + iy $ , $ y \ne 0 $ - - - - - - - (1)
Also, we are given that
 $ \Rightarrow a = {z^2} + z + 1 $ is real. - - - - - - - (2)
Putting value of equation (1) in equation (2), we get
 $ \Rightarrow a = {\left( {x + iy} \right)^2} + \left( {x + iy} \right) + 1 $
 $ \Rightarrow a = {x^2} + 2xiy + {i^2}{y^2} + x + iy + 1 $ - - - - - - - - - (3)
Now, we know that $ i = \sqrt { - 1} $
 $ \Rightarrow {i^2} = {\left( {\sqrt { - 1} } \right)^2} = - 1 $
Putting this in equation (3), we get
 $ \Rightarrow a = {x^2} - {y^2} + 2xyi + x + iy + 1 $
Now, separate the real part and imaginary part
 $ \Rightarrow a = {x^2} - {y^2} + x + 1 + 2xyi + iy $
 $ \Rightarrow a = \left( {{x^2} - {y^2} + x + 1} \right) + i\left( {2xy + y} \right) $ - - - - - - - (4)
Now, according to equation (2), the equation is real.
For a equation to be real, its imaginary part has to be zero.
Therefore, in equation (4), $ \left( {2xy + y} \right) $ should be equal to zero.
 $
   \Rightarrow \left( {2xy + y} \right) = 0 \\
   \Rightarrow y\left( {2x + 1} \right) = 0 \\
  $
But, according to equation (1) , $ y \ne 0 $ .
 $
  2x + 1 = 0 \\
  2x = - 1 \\
  x = \dfrac{{ - 1}}{2} \;
  $
Now, the equation is real. So, equation (4) will become
 $ \Rightarrow a = \left( {{x^2} - {y^2} + x + 1} \right) $
Putting the value of x in above equation
 $
   \Rightarrow a = \left( {{{\left( { - \dfrac{1}{2}} \right)}^2} - {y^2} - \dfrac{1}{2} + 1} \right) \\
   \Rightarrow a = \dfrac{1}{4} - {y^2} - \dfrac{1}{2} + 1 \\
   \Rightarrow a = \dfrac{1}{4} + \dfrac{1}{2} - {y^2} \\
   \Rightarrow a = \dfrac{3}{4} - {y^2} \;
  $
So, clearly the value of $ a $ is less than $ \dfrac{3}{4} $ as $ {y^2} $ is subtracted from it.
Out of all four options, only option d is greater than or equal to $ \dfrac{3}{4} $ .
Therefore, our answer is option d.
So, the correct answer is “Option D”.

Note: This question can also be solved using a simple method.
 $
   \Rightarrow a = {z^2} + z + 1 \\
   \Rightarrow {z^2} + z + (1 - a) = 0 \;
  $
Here, $ D < 0 $
 $
   \Rightarrow {b^2} - 4ac < 0 \\
   \Rightarrow {\left( 1 \right)^2} - 4\left( 1 \right)\left( {1 - a} \right) < 0 \\
   \Rightarrow 1 - 4 + 4a < 0 \\
   \Rightarrow 4a < 3 \\
   \Rightarrow a < \dfrac{3}{4} \;
  $
Which proves our answer above.