Algebra and Complex Numbers PDF

# Algebra ## **UNIT 1: Elementary Theorem** on the Roots of an Equation **# Quadratic Equation** * **ax² + bx + c = 0 , a ≠ 0** **-> ax² + bx = -c** **-> 4a. ax² + 4a.bx = -4ac** **-> (2ax)² + 2(2ax).b = -4ac** **Add b² both side** **-> (2ax + b)² = b² - 4ac** **-> 2ax + b = ±√(b² - 4ac)*...

# Algebra ## **UNIT 1: Elementary Theorem** on the Roots of an Equation **# Quadratic Equation** * **ax² + bx + c = 0 , a ≠ 0** **-> ax² + bx = -c** **-> 4a. ax² + 4a.bx = -4ac** **-> (2ax)² + 2(2ax).b = -4ac** **Add b² both side** **-> (2ax + b)² = b² - 4ac** **-> 2ax + b = ±√(b² - 4ac)** **-> x = -b ±√(b² - 4ac)/2a** **-> x₁ = -b +√(b² - 4ac) /2a , x₂ = -b - √(b² - 4ac) /2a** **-> x₁ + x₂ = -2b/2a = -b/a** **-> x₁. x₂ = (-b + √(b² - 4ac)/2a) . (-b - √(b² - 4ac)/2a)** **-> x₁. x₂ = (b² - (b² - 4ac))/4a² = c/a** **Let's consider a(x - x₁)(x - x₂)** **-> a[x² - x₁x₂ - x₂x₁ + x₁x₂]** **-> ax² - ax(x₁ + x₂) + ax₁x₂** **-> ax² - ax(-b/a) + ac/a** **-> ax² + bx + c = 0** **-> a(x - x₁)(x - x₂) = ax² + bx + c = 0** **① If D = 0 : b² - 4ac = 0** **-> b² = 4ac** **-> x₁ = x₂ = -b/2a i.e perfectly.** **a(x - x₁)(x - x₂) = a(x - x₁) 2** **② If D > 0: roots are real & distinct** **③ If D < 0 : roots don't exist.** **x² - 2x = 0** **-> x(x - 2) = 0** **-> x = 0 , x = 2** **x² - 2x - 1 = 0** **D = 4 - 4 x 1 x (-1) at c = -1** **-> D = 4 + 4 = 8** **-> x = 2 ± √(4 + 4 x 1 x 1)/2 x 1** **-> 2[ 1 ± √2]** **-> x² - x + 1 + i = 0** * **o if a , b , ac are real then roots are complex conjugate** **o if a, b, ac are not real then roots are not real.** **# Polynomials f(x) = axⁿ + bxⁿ⁻¹ +--- + c₁x + c₀** **-+---** **many terms** **is called a polynomial** **# Remainder Theorem** **-> f(x) = (x - a)q(x) + r** **Divident divisor Remainder** **[R = f(a)]** **Case **If a polynomial f(x) is divided by (x - a) then a remainder (x - a) is a remainder **If R = 0, this implies this division is exact then f(x) = (x - a)q(x)** **(x - a) is a factor of f(x)** ## **# Without division find (R)**. **x²-3x² - x - 6 is divided x + 3** **f(-3) = 81 - 3 x 9 + 3 - 6** **= 81 - 27 - 3 = 54 - 3 = (51)** ## **# Factor Theorem** **If f(a) = 0** **-> f(a) = 0 then the polynomial f(x) has ( x - a )** **# Theorem of root of Equation** **-> It is an alternate method to the usual division method but it is only applicable when the division is linear. Suppose the quotient is q(x) and remainder is r.** **① f(x) = 2x³ + 3x²-2x - 5** **-> x = 1 3x -2x -5 g(x) = x - a** **-> 2x³ 2x³ + 3x² -2x - 5** **-> -5x³ + 2x² -5x² -2x - 5** **-> 5x³ 5x² 2x** **-> 10x² 20x** **-> 20x - 2x - 5** **-> 18x² - 5** **# f(x) = q(x) + r** **-> f(x) = x³ +3x² -2x - 5** **1 3 0 -2 -5** **x² 2x² 10x² 20x² 36** **1 5 10 18 31** **f(x) = 1x³ + 5x² + 10x + 18** **② x³+3x²-2x-5 (x - 2)** **1 3 -2 -5** **x² 2x² 10x² 16** **1 5 8 11** **f(x) = x² + 5x + 8 + (11) / (x - 2)** **q(x)** **③ 2x⁵ - x³ + 2x -1 / x+2** **0 - 1 0 2 -1** **22⁵ - 4x⁴ - 2 78 - 14 38 - 60** **2x⁴ - 4x³ + 7x² - 14x + 30 = 61** **q(x)** **f(x) = (x - 2)(x² + 5x + 8) + 11** ## **# Factor Form of a Polynomial:** Let f(x) which is equal to c₀xⁿ + c₁xⁿ⁻¹ +--- + cₙ₋₁x + cₙ where c₀ ≠ 0, is a degree n polynomial. **If f(a) = 0:** has a root a, then (x - a) is a factor of f(x) by factor theorem. **f(x) = (x - a₁) Q(x)** **If Q(x) = 0 : Has a root a₂ then (x - a₂) is a factor of Q(x).** **f(x) = (x - a₁)(x - a₂) Q₁(x)** **f(x) = (x - a₁)(x - a₂)----(x - aₙ) Constant** ## **# Theorem on Roots of Equation** **# Factored form of a polynomial of degree 'n':** - f(x) = a₀xⁿ + a₁xⁿ⁻¹ +--- + aₙ₋₁x + aₙ where a₀≠ 0 **-> f(x) = 0** **(Polynomial)** **If 'α' be the root of poly. equation:** **f(x) = (x - α₁) q(x)** **where q(x) is one degree less.** **If 'α₂' be the root of q(x)** **-> q(x) = 0** **-> q(x) = (x - α₂) q₁(x)** **-> f(x) = (x - α₁)(x - α₂)----(x - αₙ) aₙ** **# Let f(x) = 0 be an equation where f(x) = a₀xⁿ + a₁xⁿ⁻¹ +--- + aₙ₋₁x + aₙ with degree 'n', x₁----- αₙ will be the 'n' distinct roots of the equation.** **-> f(x) = 0** **-> f(α₁) = (α₁ - α₁)q(α₁) = 0** **If α₂ α₁ be another root** **-> f(α₂) = (α₂ - α₁)q(α₂) = 0** **-> α₂ - α₁ ≠ 0** **-> q(α₂) = 0** ## **# Fundamental Theorem of Algebra:** Every algebraic equation with complex coefficient has a complex root (it may be real or imaginary). **NOTE:** Every equation of degree 'n' has exactly 'n' roots. If a root of multiplicity 'm' be counted as 'm' roots (repeated roots). **Every polynomial or Integral Ration equation of degree 'n' is a product of 'n' linear factors/ 'n' distinct factors.** **-> (x - a₁)^m₁(x - a₂)^m₂----(x - aₙ)^mₙ** **# Relationship b/w Roots & Coef of a Equation** * **f(x) = a₀xⁿ + a₁xⁿ⁻¹ +--- + aₙ is a polynomial equation** * **If f(α) = 0 , polynomial equation** **-> α [f(α) = 0] , α is a root/solution of equation** * **x²-3x+2=0** **-> x²-2x-x+2 = 0** **-> x(x-2) -1(x-2) = 0** **-> (x-2)(x-1)** **-> x = 2, 1** **-> Coef. -> 1 , -3 , 2** **-> a₀ = 1 - 3 + 2 = 0** **-> Then (x-1) is a root of f(x) equation** * **f(2) = 4-6+2 = 0** **-> (x - 2) is a root of equation** **-> a₀xⁿ + a₁xⁿ⁻¹ + a₀ general form of n° equation** **Let a₁, a₂ be the root.** **-> x² + a₀x + a₀ = 0** **-> (x-a₁)(x-a₂) = 0** **-> x²-a₁x-a₂x+ a₁a₂ = 0** **-> x² - (a₁ + a₂)x + a₁a₂ = 0** **On comparing** * **-a₁ = a₁ + a₂ ** **-> a₁ + a₂ = -a₁/ a₀** * **a₁a₂ = a₂/a₀** **By condition that ax² + bx + c = 0 have double root.** **-> a₀ = -b + c** **sum: x + x = 2x = -b /a** **product: x² = c /a** **Equation: x² - 2ax + a² = 0** ## **RESULT:** a₀xⁿ+ a₁xⁿ⁻¹ +--- + aₙ = 0 **If the root with the form of 'α' then divides a₀ & aₙ then 'α' will be the root of the equation** ## **Q3** **x³ - 9x² + 23x - 15 = 0** **Given roots are in the form of A.P.** **Let a-d, a, a+d be the roots of the given equation.** **a - d + a + a + d = 9** **-> 3a = 9 -> a = 3** **x³ + bx² + rx = +23** **-> (a - d)a + a(a + d) + (a + d)(a - d) = 23** **-> a²-ad + a²+ad + a² - ad +ad - d² = 23** **-> 3a²-d² = 23** **-> 3 x 9 - d² = 23** **-> d² = 23 - 27 = -4 -> d = ±2** ## **Q8** **8x² - 14x² + x - 1 = 0 all in G.P.** **Let a, ar, ar² are in G.P.** **-> ax x ar x ar² = 1/8** **-> a³r³ = 1/8** **-> a. r = 1/2 -> a = 1/2r** **a + ar + ar² = 14** **-> 1/2r + 1/2r³ = 14** **-> 1 + r² = 28** **-> r² = 27** ## **Q4** **a₀x⁴ + a₁x³ + a₂x² + a₃x + a₄ = 0** **Let α, β, γ, δ be the roots.** **-> ∑ = α + β + γ + δ = -a₁/ a₀ = -b** **-> ∑₂ = αβ + βγ + γδ + δα + βδ + αγ = a₂/ a₀ = c** **-> (α + β)(γ + δ) + αγ + βδ = c/a₀** **-> ∑₃ = αβγ + βγδ + αγδ + αβδ = -a₃/ a₀ = d** **-> αβ(γ + δ) + δd(α + β) = - a₃/ a₀** **-> ∑₄ = αβγδ = a₄/ a₀ = e** ## **Formation of equation when roots are given:** **Case 1:** If all roots are rational **Case 2:** Some roots are irrational **Case 3:** roots are complex **Que** Find an equation of lowest degree with the rational coefficient. **① Two of the roots are 2 & 3** **② Three .... 1, 2, 3** **-> (x - 2)(x - 3) = 0** **-> x² - 5x + 6 = 0** **-> (x - 1)(x - 2)(x - 3)** **-> x³ - 6x² + 11x - 6 = 0** **Equation with all rational roots a₁, a₂-----aₙ is given: (x - a₁)(x - a₂)----(x - aₙ) = 0** ## **Que** **Theorem:** Irrational roots occur in pair. **For example:** x² + x + √3 **x² - x -√3** The irrational roots of an equation with rational coef. occurred in pair i.e. **If an equation a₀xⁿ + a₁xⁿ⁻¹ +--- + aₙ₋₁x + aₙ = 0 where a₀ ≠ 0 & a₁ to aₙ has an irrational root a+√b then it also has an irrational root a - √b** **-> a + √b -> a + √b** **-> -a + √b** **-> -a - √b** **Que** Find an equation with rational coef. which has 2 and 2+√3 as two of its roots. **-> (x - 2)(x - (2 + √3))(x - (2 - √3)) = 0** **-> (x - 2)(x - 2 - √3)(x - 2 + √3) = 0** **-> [x² - 4x - √3x + 4 + 2√3][x - 2 + √3] = 0** **-> x³ - 2x² + 8x = 0** **Theorem:** Some complex roots occur in the conjugate pairs. **a₀xⁿ + a₁xⁿ⁻¹ +--- + aₙ₋₁x + aₙ = 0** **When a₀ ≠ 0 & a₁ to aₙ has complex root 'a + ib' then it also has complex root 'a - ib'.** **-> Complex roots of an equation with real coef. always occur in conjugate pairs.** **Remark:** The complex roots of an equation with complex no. coef. may or may not occur in complex conjugate pairs. **For example:** x² + 7ix + 12 = 0 **-> x² - 3ix - 4x -12 = 0** **-> x(x - 3i) - 4(x - 3i) ** **-> (x - 3i)(x - 4) = 0** **Que** Find the set (1 + 3i)(1 - 3i) find the equation. **-> [x - (1 + 3i)][x - (1 - 3i)] = 0** **-> (x - 1 - 3i)(x - 1 + 3i) = 0** **-> x² - 2x + 3ix - x + 1 - 3i - 3ix + 3i + 9i² = 0** **-> x² - 2x + 10 = 0** ## **# Upper Limit to Real Roots of Equation:** **Any number which exceed all real roots of a real equation is called upper limit to the real roots of that equation.** **-> Suppose α β are the roots, then 'γ' is the upper limit of the roots.** **-> x² + 4x + 8 = 0** **-> x² + 5x + 6 = 0** **-> x = -3, -2.** **Any true number is the upper limit to the real numbers of an equation having no negative coefficient.** ## **# Upper Limit to Real Roots of Equation:** **Any number which exceed all real roots of a real equation is called upper limit to the real roots of that equation.** **-> Suppose α β are the roots, then 'γ' is the upper limit of the roots.** **-> x² + 4x + 8 = 0** **-> x² + 5x + 6 = 0** **-> x = -3, -2.** **Any true number is the upper limit to the real numbers of an equation having no negative coefficient.** ## **# Theorem:** **If in a real equation, f(x) = a₀xⁿ + a₁xⁿ⁻¹ +--- + aₙ₋₁x + aₙ = 0** **The first +ve coeff. is preceded by "k" coefficient which are all zero.** **If 'G' denotes the greatest of the numerical value of the -ve coefficient, then each real root is less than 1 + k√(G/a₀) -> 1 + k√(G/a₀).** **Q/ x⁵ + 4x² - 7x² - 40x + 1 = 0** **-> x⁵ + 4x² + 0x³ -7x² - 40x + 1 = 0** **k = 3, G = 40** **1 + 3√(40/1)** **-> x³ + 8x² - 9x + c² = 0** **k = 2, G = 9** **1 + 2√(9/1)** **-> 1 + 3 = 4** **Theorem: If in a real algebraic equation in which the coefficient of the highest power of the unknown is +ve, then the sum of all the +ve coefficients divided by the greatest quotient as obtained by unit in the u.l. of the roots.** **-> a₀xⁿ + a₁xⁿ⁻¹ + --- + aₙ₋₁x + aₙ** **① a₀x⁰ , If -ve multiply by (-1)** **-> a₀ + a₁** **② a₀ + a₁ -> a₀** **③ 9 + 1 = 9** **Que** x³ + 8x² -9x + c² = 0 **① 90 = 1** **② 9 = 9** **③ 1 + 8 = 9** **-> 1 + 1 = 2 u.l of equation** **18 - 9 - 24 31 +1** **# Newton's Method for Integral Roots** **f(x) = a₀xⁿ + a₁xⁿ⁻¹ +--- + aₙ₋₂x² + aₙ₋₁x + aₙ = 0** **-> aₙ₋₁x + aₙ = -a₀xⁿ - a₁xⁿ⁻¹ - a₂xⁿ⁻² ---- aₙ₋₂x** **-> aₙ₋₁x + aₙ = x²[-a₀xⁿ⁻² - a₁xⁿ⁻³ ---- aₙ₋₂] ** **-> x² divides aₙ₋₁x + aₙ** **-> x divides aₙ₋₁x + aₙ/ x** **-> aₙ₋₂x² + aₙ₋₁x + aₙ = x²[-a₀xⁿ⁻⁴ - a₁xⁿ⁻⁵ ---- aₙ₋₃] ** **-> x² divides aₙ₋₂x² + aₙ₋₁x + aₙ** **-> x divides aₙ₋₂x² + aₙ₋₁x + aₙ/ x** **-> aₙ₋₃x³ + aₙ₋₂x² + aₙ₋₁x + aₙ** **-> x divides aₙ₋₃x³ + aₙ₋₂x² + aₙ₋₁x + aₙ/ x** **q. f(x) = x⁴ - 9x³ + 24x² -23x + 15 = 0** **f(3) = 81 - 9 x 27 + 24 x 9 - 23 x 3 + 15** **-> 3 - 23 +15** **-> 3** **-> - 23 + 5 = - 18** **-> - 18 divided by 3** **-> 3/ 24 - 6 = 18** **-> 3/ - 9 + 6 = - 3** **-> 3/ 0 - 1 + 1 = 0** **Thus f(x) is divisible by 3.** **Que** x⁴ - 9x³ + 24x² -23x + 15 **1 -9 24 -23 15 ** **-2 -6 6 -5** **0 - 3 18 - 18** **-> '3' is a root of poly equation** **1 -9 24 -23 15 ** **-5 -4 -20 - 3** **-> '5' is the root of equation** **1 - 9 24 -23 15** **-3 -5 -28** **x⁴ - 8x³ - 21x² + 22x + 40 = 0** **① 40 = ±1 , ±2 , ±4, ±5 , ±8 ,±10, ±20, ±40** **② 2** **-> Out apply u.l.** **-> -21 -22 -21 40** **-> - 2 is the u.l.** **-> k = 1, G = 21** ** -> ** 1 + √(21/1) = 1 + √21 = 5 ** **-> E2, 5 -> at x = 4** **1 - 2 -21 22 40 ** **4 8 -10 32** **-> 13 32** **-> so 4 is not the root** ** -> at x = 2** **1 -2 -21 22 40 ** **2 0 21 20** **-> 0 2 0 42** **-> '2' is the root of equation** # **Complex Numbers:** **-> z = x + iy** **-> z = r(cosθ + isinθ)** **x = r cosθ** **-> y = r sinθ** **r² + y² = r²(cos²θ + sin²θ)** **-> r² + y² = r² = √(x² + y²)** **-> sinθ = y/r** **Complex Numbers are an extension of real numbers, and they come into the existence to get √(-1) = i.** # **General Complex Number is z = x + iy, where x & y are the real numbers & i = √-1.** **A set of complex numbers is denoted by "C". If it is in the form c = x + iy , x & y ∈ R & i = √-1. ** **A complex number is '0' iff x & y are '0'.** **-> z = x - iy** **-> |z| = √(x² + y²)** **-> z = x + + iy** **-> |z| = √(x² + y²) -> |z| = r** **-> θ is the angle between 0Z with the x-axis in anticlockwise direction.** **-> Then y = r sinθ = tanθ -> θ = tan⁻¹(y/x)** ## **Q1** **-> 1 + √3i** **x = 1, y = √3** **-> θ = tan⁻¹(√3/1) = π/3** **-> r = √(x² + y²) = √(1 + 3) = √4 = 2** **-> z = r(cosθ + isinθ)** **-> z = 2(cosπ/3 + isinπ/3)** ## **Q2** **-> z = -1 - √3i** **-> θ = tan⁻¹(√3/1) = π/3** **-> (-1, -√3)** **-> z = r(cosθ + isinθ)** **-> z = 2(cos4π/3 + isin4π/3)** **NOTE:** **-> tan⁻¹(y/x) if x, y > 0** **-> π + tan⁻¹(y/x) if x < 0 , y > 0** **-> π + tan⁻¹(y/x) if x < 0 , y < 0** **-> 2π + tan⁻¹(y/x) if x > 0 , y < 0** **-> 2 = r(cosθ + isinθ)** **-> where θ = cos⁻¹(x/r) , sin⁻¹(y/r)** **Que** Find polar representation of 2 where 2 = 1 + cosα **-> x = 1 + cosα** **-> y = sinα** **-> r = √((1 + cosa)² + sin²α) = √(1 + 2 cosα + cos²α + sin²α) = √(2 + 2 cosα)** **-> r = √(2(1 + cosα)) = √(2 cos²(α/2)) = √2|cosα/2|** **-> θ = tan⁻¹(sinα/ (1 + cosα))** **-> θ = tan⁻¹(2 sin(α/2) cos(α/2)/ 2 cos²(α/2))** **-> θ = tan⁻¹(tan(α/2)) ** **-> θ = α/2** ## **De' Moivre's Theorem for n ∈ N:** **(cosθ + isinθ)ⁿ = cos nθ + i sin nθ ; n ∈ N** **Base Condition:** When n = 1 **-> (cosθ + isinθ)¹ = cosθ + isinθ** **LHS = RHS** **-> Result is true for n = 1** **Let the result be true for n = k** **-> (cosθ + isinθ)ᵏ = cos kθ + isin kθ (1)** **We will prove that the result is true for n = k + 1.** **-> (cosθ + isinθ) ᵏ⁺¹ = cos(k + 1) θ + i sin (k + 1) θ** **-> (cosθ + isinθ) ᵏ⁺¹=(cosθ + isinθ) ᵏ (cosθ + isinθ)** **-> (cosθ + isinθ) ᵏ⁺¹ = (cos kθ + isin kθ)(cos θ + isin θ)** **-> cos kθ cosθ + i cos kθ sin θ+ isin kθ cosθ - sin kθ sin θ** **-> (cos kθ cosθ - sin kθ sin θ) + i (cos kθ sin θ + sin kθ cos θ)** **-> cos (k + 1) θ + i sin(k + 1)θ** **-> RHS** **Hence, the result is true for n = k + 1** **-> (cosθ + isinθ)ⁿ = cos nθ + isin nθ** ## **De' Moivre's Theorem for integers:** **(cos θ + isin θ)ⁿ = cos nθ + i sin nθ** **Case 1: n > 0** **-> n is a N** **Case 2: n = 0** **-> n = 0** **Case 3: n < 0** **-> n = -m** **-> m > 0** **-> (cos θ + isin θ)ⁿ** **-> (cos θ + isin θ)⁻ᵐ** **-> (cos θ + isin θ)⁻¹ ᵐ** **-> 1 / (cosθ + isinθ) ᵐ** **-> (cos mθ + i sin mθ)⁻¹ [Rationalising]** **-> (cos mθ - i sin mθ)/ (cos²mθ + sin²mθ)** **-> cos(-m)θ + i sin(-m) θ = (cos nθ + i sin nθ)** **-> RHS.** ## **Questions** **Que 1** Find the |z| and arg(z) of (1 - i)¹⁰ (√3 + i)⁵ **-> x + iy -> x = rcosθ = 1** **-> y = rsinθ = -1** **|z| = √(x² + y²) -> |z| = √(1² + 1²) = √2** **-> θ = tan⁻¹(-1/1) = -π/4** **-> θ = 2π - π/4 = 7π/4** **-> z = (1 - i)¹⁰** **-> z = [(√2)(cos ( - π/4) + isin (-π/4))] ¹⁰ ** **-> z = (√2) ¹⁰ [cos (- 5π/2 ) + i sin (-5π/2)]** **-> θ = ( - 5π/2) = ( - 11π/2)** **-> z = [ (√2)¹⁰ (cos (- 11π / 2) + isin (-11π/ 2))]** **-> z = [(√2)¹⁰ (cos (- 11π / 2) + isin (- 11π/ 2))]** **-> z = [ (√2)¹⁰ (cos (π/2) + isin (π/2))]** ** -> z₂ = [ (√3 + i)⁵ = [2(cosπ/6 + isinπ/6)] ⁵** **-> z₂ = 32 [cos5π/6 + isin 5π/6]** **z₁ z₂ = (1 - i)¹⁰ (√3 + i)⁵** **-> z₁z₂ = [(√2)¹⁰ (cos (π/2) + isin (π/2))] * 32 [ cos 5π/6 + isin 5π/6]** **-> z₁z₂ = (√2)¹⁰ * 32 [cos((5π/6) + (π/2)) + isin((5π/6) + (π/2)) ]** **-> z₁z₂ = (√2)¹⁰ * 32 [cos(4π/3) + isin (4π/3)]** **-> z₁z₂ = (√2)¹⁰ * 32 [cos(8π/3) + isin (8π/3)]** **-> z₁z₂ = (√2)¹⁰ * 32 [cos(2π/3) + isin (2π/3)]** **-> z₁z₂ = 1024 [cos(2π/3) + isin (2π/3)]** **-> z₁z₂ = 1024 * 1/2 * (- 1 + √3i)** **-> z₁z₂ = 512(-1 + √3i)** # **When we multiply two polar representations, then their arg get add & when we divide then arg get substracted.** **-> z₁z₂ = cos θ₁ + i sin θ₁** **-> z₂ = cos θ₂ + i sin θ₂** **-> 1/z₂ = 1/(cos θ₂ + i sin θ₂) = (cos θ₂ - i sin θ₂)/ (cos²θ₂ + sin²θ₂)** **-> 1/z₂ = cos θ₂ - i sin θ₂** **-> z₁/z₂ = (cos θ₁ + i sin θ₁)(cos θ₂ - i sin θ₂) **-> z₁/z₂ = (cos θ₁ cos θ₂ - sin θ₁sin θ₂ ) + i (sinθ₁ cos θ₂ + cosθ₁ sin θ₂) ** **-> z₁/z₂ = cos (θ₁ + θ₂) + i sin ( θ₁ + θ₂)** **-> (cos θ₁ + i sin θ₁)(cos θ₂ + i sin θ₂) --- (cos θₙ + i sin θₙ)** **-> (cos θ₁ + i sin θ₁)(cos θ₂ - i sin θ₂) = cos (θ₁ + θ₂ - θ₃ ---- θₙ) + i sin (θ₁ + θ₂ - θ₃ ---- θₙ)** **-> (cos θ - i sin θ)ⁿ = cos nθ - i sin nθ** ## **# Express in the form of