SATHEE: Complex Numbers 2 Question 9

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) $\frac{1}{3}$

(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 $α, β$ , then

$α, β = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$

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.

$\Rightarrow y \neq 0$

Method I Let $z = x + i y$

$\begin{array}{lc} ∴ & a = (x + i y)^{2} + (x + i y) + 1 \\ \Rightarrow & (x^{2} - y^{2} + x + 1 - a) + i (2 x y + y) = 0 \\ \Rightarrow & (x^{2} - y^{2} + x + 1 - a) + i y (2 x + 1) = 0 \end{array}$

It is purely real, if $y (2 x + 1) = 0$

but imaginary part of $z$ , i.e. $y$ is non-zero.

$\Rightarrow 2 x + 1 = 0$ or $x = - 1 / 2$

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

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

Method II Here, $z^{2} + z + (1 - a) = 0$

$\begin{array}{ll} ∴ & z = \frac{- 1 \pm \sqrt{1 - 4 (1 - a)}}{2 \times 1} \\ \Rightarrow & z = \frac{- 1 \pm \sqrt{4 a - 3}}{2} \end{array}$

For $z$ do not have real roots, $4 a - 3 < 0 \Rightarrow a < \frac{3}{4}$

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

Answer:

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Complex Numbers 2 Question 9

9. Let z be a complex number such that the imaginary part of z is non-zero and a=z2+z+1 is real. Then, a cannot take the value

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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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