Complex Numbers 2 Question 9
9. 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
(a) -1
(b) $\frac{1}{3}$
(c) $\frac{1}{2}$
(d) $\frac{3}{4}$
(2012)
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Answer:
Correct Answer: 9. (d)
Solution:
- PLAN If $a x^{2}+b x+c=0$ has roots $\alpha, \beta$, then
$$ \alpha, \beta=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} $$
For roots to be real $b^{2}-4 a c \geq 0$.
Description of Situation As imaginary part of $z=x+i y$ is non-zero.
$\Rightarrow \quad y \neq 0$
Method I Let $z=x+i y$
$$ \begin{array}{lc} \therefore & a=(x+i y)^{2}+(x+i y)+1 \\ \Rightarrow & \left(x^{2}-y^{2}+x+1-a\right)+i(2 x y+y)=0 \\ \Rightarrow & \left(x^{2}-y^{2}+x+1-a\right)+i y(2 x+1)=0 \end{array} $$
It is purely real, if $y(2 x+1)=0$
but imaginary part of $z$, i.e. $y$ is non-zero.
$\Rightarrow \quad 2 x+1=0$ or $x=-1 / 2$
From Eq. (i), $\frac{1}{4}-y^{2}-\frac{1}{2}+1-a=0$
$\Rightarrow \quad a=-y^{2}+\frac{3}{4} \Rightarrow a<\frac{3}{4}$
Method II Here, $z^{2}+z+(1-a)=0$
$$ \begin{array}{ll} \therefore & z=\frac{-1 \pm \sqrt{1-4(1-a)}}{2 \times 1} \\ \Rightarrow & z=\frac{-1 \pm \sqrt{4 a-3}}{2} \end{array} $$
For $z$ do not have real roots, $\quad 4 a-3<0 \quad \Rightarrow \quad a<\frac{3}{4}$
