Math Notes For Class 11 Chapter 3 Ex 3.2 Solved & Quiz

Question 1 (Part i): Find the complex cube roots of $8$.

Solution:
Let $x = (8)^{1/3} \implies x^3 = 8 \implies x^3 - 8 = 0$.
Using the algebraic identity $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$: $$(x - 2)(x^2 + 2x + 4) = 0$$ Case 1: $x - 2 = 0 \implies x = 2$.
Case 2: $x^2 + 2x + 4 = 0$. Applying the quadratic formula with $a=1, b=2, c=4$: $$x = \frac{-2 \pm \sqrt{2^2 - 4(1)(4)}}{2(1)} = \frac{-2 \pm \sqrt{4 - 16}}{2} = \frac{-2 \pm \sqrt{-12}}{2}$$ $$x = \frac{-2 \pm 2\sqrt{3}i}{2} = 2 \left( \frac{-1 \pm \sqrt{3}i}{2} \right)$$ Substituting $\omega = \frac{-1 + \sqrt{3}i}{2}$ and $\omega^2 = \frac{-1 - \sqrt{3}i}{2}$: $$x = 2\omega \quad \text{and} \quad x = 2\omega^2$$
Answer: The cube roots of $8$ are $$\{2, 2\omega, 2\omega^2\}$$

Question 1 (Part ii): Find the complex cube roots of $-27$.

Solution:
Let $x = (-27)^{1/3} \implies x^3 = -27 \implies x^3 + 27 = 0$.
Using $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$: $$(x + 3)(x^2 - 3x + 9) = 0$$ Case 1: $x + 3 = 0 \implies x = -3$.
Case 2: $x^2 - 3x + 9 = 0$. Applying the quadratic formula ($a=1, b=-3, c=9$): $$x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(9)}}{2(1)} = \frac{3 \pm \sqrt{9 - 36}}{2} = \frac{3 \pm \sqrt{-27}}{2}$$ $$x = \frac{3 \pm 3\sqrt{3}i}{2} = -3 \left( \frac{-1 \mp \sqrt{3}i}{2} \right)$$ Thus, $x = -3\omega^2$ and $x = -3\omega$.
Answer: The cube roots of $-27$ are $$\{-3, -3\omega, -3\omega^2\}$$

Question 1 (Part iii): Find the complex cube roots of $64$.

Solution:
Let $x = (64)^{1/3} \implies x^3 - 64 = 0 \implies (x - 4)(x^2 + 4x + 16) = 0$.
Case 1: $x - 4 = 0 \implies x = 4$.
Case 2: $x^2 + 4x + 16 = 0$: $$x = \frac{-4 \pm \sqrt{16 - 64}}{2} = \frac{-4 \pm \sqrt{-48}}{2} = \frac{-4 \pm 4\sqrt{3}i}{2} = 4 \left( \frac{-1 \pm \sqrt{3}i}{2} \right)$$ Hence, $x = 4\omega$ and $x = 4\omega^2$.
Answer: The cube roots of $64$ are $$\{4, 4\omega, 4\omega^2\}$$

Question 2 (Part i): Evaluate $(1 - \omega - \omega^2)^7$.

Solution:
Since $1 + \omega + \omega^2 = 0 \implies -\omega - \omega^2 = 1$.
Substitute this into the expression: $$(1 - \omega - \omega^2)^7 = [1 + (-\omega - \omega^2)]^7 = (1 + 1)^7 = 2^7$$ $$2^7 = 128$$
Answer: $$128$$

Question 2 (Part ii): Evaluate $(1 - 3\omega - 3\omega^2)^5$.

Solution:
Factor out $-3$ from the last two terms: $$(1 - 3\omega - 3\omega^2)^5 = [1 - 3(\omega + \omega^2)]^5$$ Since $\omega + \omega^2 = -1$: $$= [1 - 3(-1)]^5 = (1 + 3)^5 = 4^5$$ $$4^5 = 1024$$
Answer: $$1024$$

Question 2 (Part iii): Evaluate $(1 + \omega - \omega^2)(1 - \omega + \omega^2)$.

Solution:
Using $1 + \omega = -\omega^2$ and $1 + \omega^2 = -\omega$:
First factor: $(1 + \omega - \omega^2) = (-\omega^2 - \omega^2) = -2\omega^2$.
Second factor: $(1 - \omega + \omega^2) = (1 + \omega^2 - \omega) = (-\omega - \omega) = -2\omega$.
Multiply both terms: $$(-2\omega^2)(-2\omega) = 4\omega^3$$ Since $\omega^3 = 1$: $$4(1) = 4$$
Answer: $$4$$

Question 3 (Part i): Prove that $x^3 - y^3 = (x - y)(x - \omega y)(x - \omega^2 y)$.

Solution:
Consider the Right Hand Side (RHS): $$\text{RHS} = (x - y) [(x - \omega y)(x - \omega^2 y)]$$ Expand the inner product: $$(x - \omega y)(x - \omega^2 y) = x^2 - \omega^2 xy - \omega xy + \omega^3 y^2$$ $$= x^2 - (\omega + \omega^2)xy + (1)y^2$$ Since $\omega + \omega^2 = -1$: $$= x^2 - (-1)xy + y^2 = x^2 + xy + y^2$$ Multiply by $(x - y)$: $$\text{RHS} = (x - y)(x^2 + xy + y^2) = x^3 - y^3 = \text{LHS}$$
Answer: $$\text{Proved that } x^3 - y^3 = (x - y)(x - \omega y)(x - \omega^2 y)$$

Question 3 (Part ii): Prove that $x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x + \omega y + \omega^2 z)(x + \omega^2 y + \omega z)$.

Solution:
Expand the second and third factors of the Right Hand Side (RHS): $$(x + \omega y + \omega^2 z)(x + \omega^2 y + \omega z)$$ $$= x^2 + \omega^2 xy + \omega xz + \omega xy + \omega^3 y^2 + \omega^2 yz + \omega^2 xz + \omega^4 yz + \omega^3 z^2$$ Group terms with common algebraic factors, using $\omega^3 = 1$ and $\omega^4 = \omega$: $$= x^2 + y^2(1) + z^2(1) + xy(\omega + \omega^2) + yz(\omega^2 + \omega) + xz(\omega + \omega^2)$$ Substitute $\omega + \omega^2 = -1$: $$= x^2 + y^2 + z^2 - xy - yz - zx$$ Now multiply by the first factor $(x + y + z)$: $$\text{RHS} = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx) = x^3 + y^3 + z^3 - 3xyz = \text{LHS}$$
Answer: $$\text{Proved that } x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x + \omega y + \omega^2 z)(x + \omega^2 y + \omega z)$$

Question 4 (Part i): Prove that $(-1 + \sqrt{-3})^6 + (-1 - \sqrt{-3})^6 = 128$.

Solution:
Recall that $\omega = \frac{-1 + \sqrt{-3}}{2} \implies -1 + \sqrt{-3} = 2\omega$.
Similarly, $\omega^2 = \frac{-1 - \sqrt{-3}}{2} \implies -1 - \sqrt{-3} = 2\omega^2$.
Substitute these into the Left Hand Side (LHS): $$\text{LHS} = (2\omega)^6 + (2\omega^2)^6$$ $$= 2^6 \cdot \omega^6 + 2^6 \cdot \omega^{12} = 64(\omega^3)^2 + 64(\omega^3)^4$$ Since $\omega^3 = 1$: $$\text{LHS} = 64(1)^2 + 64(1)^4 = 64 + 64 = 128 = \text{RHS}$$
Answer: $$\text{Proved that } (-1 + \sqrt{-3})^6 + (-1 - \sqrt{-3})^6 = 128$$

Question 4 (Part ii): Prove that $(1 + \omega)(1 + \omega^2)(1 + \omega^4)(1 + \omega^8) = 1$.

Solution:
Reduce high powers of $\omega$: $$\omega^4 = \omega^3 \cdot \omega = (1)\omega = \omega$$ $$\omega^8 = (\omega^3)^2 \cdot \omega^2 = (1)^2 \omega^2 = \omega^2$$ Substitute these back into the LHS: $$\text{LHS} = (1 + \omega)(1 + \omega^2)(1 + \omega)(1 + \omega^2) = [(1 + \omega)(1 + \omega^2)]^2$$ Since $1 + \omega = -\omega^2$ and $1 + \omega^2 = -\omega$: $$(1 + \omega)(1 + \omega^2) = (-\omega^2)(-\omega) = \omega^3 = 1$$ $$\text{LHS} = (1)^2 = 1 = \text{RHS}$$
Answer: $$\text{Proved that } (1 + \omega)(1 + \omega^2)(1 + \omega^4)(1 + \omega^8) = 1$$

Question 5 (Part i): Find the four fourth roots of $16$.

Solution:
Let $x = (16)^{1/4} \implies x^4 = 16 \implies x^4 - 16 = 0$.
Factor using $a^2 - b^2 = (a - b)(a + b)$: $$(x^2 - 4)(x^2 + 4) = 0$$ Case 1: $x^2 - 4 = 0 \implies x^2 = 4 \implies x = \pm 2$.
Case 2: $x^2 + 4 = 0 \implies x^2 = -4 \implies x = \pm \sqrt{-4} = \pm 2i$.
Answer: The four fourth roots of $16$ are $$\{2, -2, 2i, -2i\}$$

Question 5 (Part ii