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Question: Let z be a complex number such that the imaginary part of z is nonzero and \( a = {z^2} + z + 1 \) i...

Let z be a complex number such that the imaginary part of z is nonzero and a=z2+z+1a = {z^2} + z + 1 is real. Then a cannot take the value:
A)−1 B)13 C)12 D)34  A) - 1 \\\ B)\dfrac{1}{3} \\\ C)\dfrac{1}{2} \\\ D)\dfrac{3}{4} \\\

Explanation

Solution

Hint : In this question, first of all we are going to put the value of z in the equation a=z2+z+1a = {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.
⇒z=x+iy\Rightarrow z = x + iy , y≠0y \ne 0 - - - - - - - (1)
Also, we are given that
⇒a=z2+z+1\Rightarrow a = {z^2} + z + 1 is real. - - - - - - - (2)
Putting value of equation (1) in equation (2), we get
⇒a=(x+iy)2+(x+iy)+1\Rightarrow a = {\left( {x + iy} \right)^2} + \left( {x + iy} \right) + 1
⇒a=x2+2xiy+i2y2+x+iy+1\Rightarrow a = {x^2} + 2xiy + {i^2}{y^2} + x + iy + 1 - - - - - - - - - (3)
Now, we know that i=−1i = \sqrt { - 1}
⇒i2=(−1)2=−1\Rightarrow {i^2} = {\left( {\sqrt { - 1} } \right)^2} = - 1
Putting this in equation (3), we get
⇒a=x2−y2+2xyi+x+iy+1\Rightarrow a = {x^2} - {y^2} + 2xyi + x + iy + 1
Now, separate the real part and imaginary part
⇒a=x2−y2+x+1+2xyi+iy\Rightarrow a = {x^2} - {y^2} + x + 1 + 2xyi + iy
⇒a=(x2−y2+x+1)+i(2xy+y)\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), (2xy+y)\left( {2xy + y} \right) should be equal to zero.
⇒(2xy+y)=0 ⇒y(2x+1)=0  \Rightarrow \left( {2xy + y} \right) = 0 \\\ \Rightarrow y\left( {2x + 1} \right) = 0 \\\
But, according to equation (1) , y≠0y \ne 0 .
2x+1=0 2x=−1 x=−12   2x + 1 = 0 \\\ 2x = - 1 \\\ x = \dfrac{{ - 1}}{2} \;
Now, the equation is real. So, equation (4) will become
⇒a=(x2−y2+x+1)\Rightarrow a = \left( {{x^2} - {y^2} + x + 1} \right)
Putting the value of x in above equation
⇒a=((−12)2−y2−12+1) ⇒a=14−y2−12+1 ⇒a=14+12−y2 ⇒a=34−y2   \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 aa is less than 34\dfrac{3}{4} as y2{y^2} is subtracted from it.
Out of all four options, only option d is greater than or equal to 34\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.
⇒a=z2+z+1 ⇒z2+z+(1−a)=0   \Rightarrow a = {z^2} + z + 1 \\\ \Rightarrow {z^2} + z + (1 - a) = 0 \;
Here, D<0D < 0
⇒b2−4ac<0 ⇒(1)2−4(1)(1−a)<0 ⇒1−4+4a<0 ⇒4a<3 ⇒a<34   \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.