IIT-JEE 2012 · Paper 1 · Mathematics Q45

← IIT-JEE 2012 · Paper 1

MathematicsSingle correct+3 / -1Question 45 of 60

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

−1

\dfrac13

\dfrac12

\dfrac34

Answer and worked solution

Official answer: D

Write $z=x+iy$ , $y\ne0$ .
$\operatorname{Im}(z^2+z+1)=2xy+y=y(2x+1)=0\implies x=-\tfrac12.$

Value of $a$ . With $z=-\frac12+iy$ :
$a=\left(\tfrac14-y^2\right)-\tfrac12+1=\tfrac34-y^2.$
Since $y\ne0$ , $a<\frac34$ : it can be −1, $\frac13$ or $\frac12$ but never $\frac34$ .

(D)

Topics:Algebra of complex numbers

← Previous question Next question →

IIT-JEE 2012 · Paper 1, Mathematics question 45