Discussion Forum

Let z be a complex number such that the imaginary part of z is non-zero and $a=z^{2}+z+1$  is real. Then, a cannot take the value

Answer:

Explanation:

 Concept involved if $ax^{2}+bx+c=0$ has roots $\alpha, \beta$ , then

 $\alpha, \beta=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$

 for roots to be real $b^{2}-4ac \geq 0$

 Description of Situation As imaginary

part of z=x+iy is non zero

 $\Rightarrow$    $y \neq 0$

 sol. 1 method let z=x+iy

 $\therefore$  $a=(x+iy)^{2}+(x+iy)+1$

$\Rightarrow$  $(x^{2}-y^{2}+x+1-a)+i(2xy+y)=0$

 $\Rightarrow$  $(x^{2}-y^{2}+x+1-a)+iy(2x+1)=0$......(i)

 It is purely real .If ,y(2x+1)=0

 but imagonary part of z, i.em y is non zero

 $\Rightarrow$  2x+1=0 or x= $-\frac{1}{2}$

$\therefore$  From Eq.(i) $\frac{1}{4}-y^{2}-\frac{1}{2}+1-a=0$

 $\Rightarrow$  $a=-y^{2}+\frac{3}{4} \Rightarrow a <\frac{3}{4}$

 II method  Here, $z^{2}+z+(1-a)=0$

 $\therefore$    $z=\frac{-1\pm\sqrt{1-4(1-a)}}{2\times1}$

  $\Rightarrow$  $z=\frac{-1\pm\sqrt{4a-3}}{2}$

 for    z do not have real roots.

  4a-3<0 $\Rightarrow$  a < $\frac{3}{4}$