Complex Numbers Exercise 3 Solutions | Basic Mathematics Grade XI by Sukunda Pustak Bhawan

In this page, you can find the complete solutions of the third exercise of Complex Numbers chapter from Basic Mathematics Grade XI book published and distributed by Sukunda Pustak Bhawan.

In the above-mentioned book, Complex Numbers is the 6th chapter and has four exercises only. Out of which, this is the solution of the third exercise.

Check: Basic Mathematics Grade 11 (Sukunda Publication) Guide:

Grade 11 Basic Mathematics by Sukunda Pustak Vawan Notes and Solutions | Nepal

Disclaimer:

Answers mentioned here are not solved by teachers. These are the solutions written by a student of Grade 11. Answers are all correct. However, the language or process of solving the questions might be informal and in examinations, you might have to add little more language and show more calculations than what has been done here. So, we highly encourage you to view these solutions as guide rather than just copying everything mentioned here.

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1. If $\rm \omega$ be a complex cube root of unity, show that

a) $\rm (1 + \omega - \omega^2)^3 - ( 1  - \omega + \omega^2)^3 = 0$

Solution:
Given,

$\rm = (1 + \omega - \omega^2)^3 - (1 - \omega + \omega^2)^3$

We know, 

$\rm 1 + \omega + \omega^2 = 0$

$\rm = (- \omega^2 - \omega^2)^3 - (1 + \omega^2 - \omega)^3$

$\rm = (-2 \omega ^2)^3 - (- \omega - \omega)^3$

$\rm = (-2 \omega^2)^3 - (- 2  \omega)^3$

$\rm = -8 \cdot \omega^6 + 8 \omega^3$

$\rm = - 8 \cdot 1 + 8 \cdot 1$

$\rm = 0$

b) $\rm (2 + \omega + \omega^2)^3 + (1 + \omega - \omega^2)^8 - (1 - 3 \omega + \omega^2)^4 = 1$

Solution:
Given,

$\rm = (2 + \omega + \omega^2)^3 + (1 + \omega - \omega^2) ^8 - (1 - 3 \omega + \omega^2)^4$

We know, $\rm 1 + \omega + \omega^2 = 0$

$\rm = (2 + (-1))^3 + (- \omega^2 - \omega^2)^8 - (1 + \omega^2 - 3 \omega)^4$

$\rm = (2 -1 )^3 + (- 2\omega ^2)^8 - (-\omega - 3 \omega)^4$

$\rm = (1)^3 + (-2 \omega^2)^8 -  (- 4 \omega)^4$

$\rm = 1 + 2^8 (\omega)^8 - 4^4 (\omega)^4$

$\rm = 1 + 2^8 (1) - 4^4(1)$

$\rm = 1 + 4^4 - 4^4$

$\rm = 1 + 0$

$\rm = 1$

2 a) If $\alpha = \frac{1}{2} ( - 1 + \sqrt{-3})$ and $\beta = \frac{1}{2} ( - 1 - \sqrt{-3})$, show that $\alpha^4 + \alpha^2 \beta^2 + \beta^4 = 0$.

Solution:
Given,

$\alpha = \frac{1}{2} ( - 1 + \sqrt{-3})$

$\beta = \frac{1}{2} ( - 1 - \sqrt{-3})$

Now,

$\alpha^4 + \alpha^2 \beta^2 + \beta^4 = ( \alpha^2)^2 + (\beta^2)^2 + \alpha^2 \beta^2)$

$\rm = (\alpha^2 + \beta^2)^2 - 2 \alpha^2 \beta^2 + \alpha^2 \beta^2$

$\rm = (\alpha^2 + \beta^2)^2 - (\alpha^2 \beta^2)$

$\rm = (\alpha^2 + \beta^2)^2 - (\alpha \beta)^2$

$\rm = (\alpha  + \beta + \alpha \beta)(\alpha + \beta - \alpha \beta)$ - (i)

And,

$\rm \alpha \beta = \left ( \frac{1}{2} ( - 1 + \sqrt{-3})  \right ) \left ( \frac{1}{2} ( - 1 - \sqrt{-3}) \right )$

$\rm = \frac{-1}{4} ( - 1 + \sqrt{-3})(1 + \sqrt{-3})$

$\rm = \frac{-1}{4} ( - 1  - \sqrt{-3} + \sqrt{-3} + \sqrt{-3} \sqrt{-3})$

$\rm = \frac{-1}{4} ( - 1 + (\sqrt{3}i)(\sqrt{3}i))$

$\rm = \frac{-1}{4} (-1 + 3 i^2)$

$\rm = \frac{-1}{4} ( - 1 -3)$

$\rm = \frac{-4}{4}$

$\rm = -1$

Again

$\rm \alpha + \beta = \frac{1}{2} ( - 1 + \sqrt{-3}) + \frac{1}{2} ( - 1 - \sqrt{-3})$

$\rm = \frac{1}{2} \left ( - 1 + \sqrt{-3} - 1 - \sqrt{-3} \right )$

$\rm = \frac{1}{2} \left ( -2 \right )$

$\rm  = -1$

From equation (i)

$\rm (\alpha  + \beta + \alpha \beta)(\alpha + \beta - \alpha \beta)$

$\rm = (1 + (-1)) ( 1 - (-1))$

$\rm = 0 \times (2)$

$\rm = 0$

4) If x = a + b, y = a$\omega$ + b$\omega^2$, z = a$\omega^2$ + b$\omega$, show that

i) x + y + z = 0

Solution:
Given,

$\rm x + y + z = (a + b) + a \omega + b \omega^2 + a \omega^2 + b \omega$

$\rm = a + a \omega + a \omega^2 + b + b \omega + b\omega^2$

$\rm = a(1 + \omega  +\omega^2) + b(1 + \omega + \omega^2)$

We know, $\rm (1 + \omega + \omega^2) = 0$

$\rm = a (0) + b(0)$

$\rm = 0 + 0$

$\rm = 0$

ii) xyz = $\rm a^3 + b^3$

Solution:
Given,

$\rm xyz = (a + b)(a \omega + b \omega^2) (a \omega^2 + b\omega)$

$\rm = (a +b) ( a \omega^3 + a b \omega^2 + ab \omega^4 + b \omega^3)$

$\rm = (a+b) ( a \cdot 1 + ab \omega^2 ( 1 + \omega^2) + b \cdot 1)$

$\rm = (a + b)(a + ab \omega^2 ( - \omega) + b)$

$\rm = (a + b) (a - ab \omega^3 + b)$

$\rm = (a + b)(a - ab \codt 1 + b)$

$\rm = (a + b)(a - ab + b)$

$\rm = a^3 +  b^3$

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About the Textbook:

Name: Basic Mathematics Grade XI

Author(s): D.R. Bajracharya | R.M. Shresththa | M.B. Singh | Y.R. Sthapit | B.C. Bajracharya

Publisher: Sukunda Pustak Bhawan (Bhotahity, Kathmandu)

Telephone: 5320379, 5353537

Price: 695 /- (2078 BS)

Website: Sukunda Publication – Sukunda Pustak Bhawan

Buy this book: Basic Mathematics : Grade XI – Sukunda Publication