Grade 12 Mathematics Textbook (Final) PDF

National Unity Government of Myanmar Ministry of Education Grade 12 Mathematics Textbook April, 2024 Grade 12 Mathematics...

National Unity Government of Myanmar Ministry of Education Grade 12 Mathematics Textbook April, 2024 Grade 12 Mathematics OE CHAPTER CONTENTS PAGE GM CHAPTER 1 Complex Numbers 1.1 Pure Imaginary Unit i 1 NU 1.2 Complex Number (x, y) 3 1.3 Operations on Complex Numbers 6 1.4 Trigonometric Form 8 E MO 1.5 Roots of Complex Numbers 16 G CHAPTER 2 Mathematical Induction NU 2.1 Introduction 20 2.2 Principle of Mathematical Induction 21 OE CHAPTER 3 Analytical Solid Geometry 3.1 Coordinates of a Point in Space 31 GM 3.2 Lines 34 3.3 Parallel, Skew and Perpendicular Lines 38 NU 3.4 Planes 43 3.5 Spheres 46 E MO CHAPTER 4 Vector Algebra 4.1 Vectors in Three Dimensions 51 G 4.2 Angle between Two Vectors and Scalar Product 57 NU 4.3 Area of a Parallelogram and Vector Product 61 4.4 Lines and Planes in Space 65 OE CHAPTER 5 Permutations and Combinations 5.1 Counting Principles 72 GM 5.2 Permutations 78 5.3 Combinations 81 NU 5.4 Techniques for Some Counting Problems 84 Grade 12 Mathematics CHAPTER 6 Conic Sections 6.1 Introduction 91 OE 6.2 Circles 91 GM 6.3 The Parabola 96 6.4 Rotation of Axes 111 NU CHAPTER 7 Trigonometric Functions 7.1 Graphs of Sine Functions 118 OE 7.2 Graphs of Cosine Functions 127 GM 7.3 Graphs of other Trigonometric Functions 131 7.4 Inverse of Trigonometric Functions 134 7.5 NU Differentiation of Trigonometric Functions 137 CHAPTER 8 Logarithmic and Exponential Functions OE 8.1 Logarithmic Functions 143 8.2 Differentiation of Logarithmic Functions 149 GM 8.3 Exponential Functions 154 8.4 Differentiation of Exponential Functions 163 NU CHAPTER 9 Application of Differentiation OE 9.1 Tangent Line and Normal Line 167 9.2 Linear Approximation 170 GM 9.3 Extreme Values of Functions 177 9.4 First Derivative Test and Second Derivative Test 182 NU for Local Extrema CHAPTER 10 Method of Integration E MO 10.1 Antiderivatives 198 10.2 Substitution Method 204 G 10.3 Integration by Parts 208 NU 10.4 Partial Fraction Method 210 Grade 12 Mathematics CHAPTER 11 Application of Integrals OE 11.1 Indefinite Integral 215 GM 11.2 Application of Indefinite Integral 215 11.3 Definite Integral 219 NU 11.4 Application of Definite Integral 223 E MO G NU OE GM NU E G MO NU OE GM NU  Grade 12 Mathematics Textbook Chapter 1 COMPLEX NUMBERS E In this chapter we introduce a new number system which is the extension of the real M number system. 1.1 Pure Imaginary Unit 𝐢 G U Since we have known that x 2 ≥ 0 for every real number x, the equation x 2 + 4 = 0 N (or) x 2 = −4 has no real solution. But if there is a pure imaginary unit i such that i2 = −1 with acceptance of the usual operations E on real numbers and i such as (2i)2 = 4i2 = 4(−1) = −4 and (−2i)2 = 4i2 = 4(−1) = −4 O then x 2 = −4 has two solutions 2i and −2i. M If we try to solve x 2 = −n for n > 0, like x 2 = −4, then √ni and −√ni are solutions because (±√ni)2 = ni2 = n(−1) = −n. G U N Example 1 E Solve x 2 + 2x + 5 = 0. O M Solution G Completing the square method U x 2 + 2x + 5 = 0 N x 2 + 2x = −5 E x 2 + 2x + 12 = −5 + 12 O M (x + 1)2 = −4 (x + 1)2 = 4i2 (∵ i2 = −1) G U x + 1 = ±2i N x = −1 ± 2i 1  Grade 12 Mathematics Textbook Using formula x 2 + 2x + 5 = 0 ax 2 + bx + c = 0 a = 1, b = 2, c = 5 E M −b ± √b 2 − 4ac x = 2a G −2 ± √4 − 4(1)(5) U = 2(1) N −2 ± √−16 = 2 E −2 ± √16i2 O = 2 M −2±4i = = −1 ± 2i. 2 Example 2 G U N Solve x 2 + 2x + 3 = 0 and check your answer. Solution E x 2 + 2x + 3 = 0 O x 2 + 2x = −3 M x 2 + 2x + 1 = −3 + 1 G (x + 1)2 = −2 U (x + 1)2 = 2i2 N x + 1 = ±√2i E x = −1 ± √2i. O Thus, x = −1 + √2i (or) x = −1 − √2i. M For x = −1 + √2i, G x 2 + 2x + 3 = (−1 + √2i )2 + 2(−1 + √2i) + 3 U N = 1 − 2√2i + 2i2 − 2 + 2√2i + 3 =1−2−2+3 =0. 2  Grade 12 Mathematics Textbook For x = −1 − √2i, x 2 + 2x + 3 = (−1 − √2i )2 + 2(−1 − √2i) + 3 = 1 + 2√2i + 2i2 − 2 − 2√2i + 3 E =1−2−2+3 M =0. G U Exercises 1.1 N 1. Solve the following equations. (a) x 2 − 6x + 10 = 0 (b) −2x 2 + 4x − 3 = 0 E (c) 5x 2 − 2x + 1 = 0 (d) 3x 2 + 7x + 5 = 0 O 2. Solve the following equations and check your answers. M (a) x 2 − 2x + 4 = 0 (b) x 2 − 4x + 5 = 0 G 3. Find the value of in for every positive integer n, where i2 = −1, i3 = i2 i, i4 = i2 i2 , etc. U N E 1.2 Complex Number (𝐱, 𝐲) O M As we have just seen in Section 1.1, there are numbers like x1 + y1 i and x2 + y2 i G such that (x1 + y1 i) + (x2 + y2 i) = (x1 + x2 ) + (y1 + y2 )i and U = x1 x2 + (x1 y2 + y1 x2 )i + y1 y2 i2 N (x1 + y1 i)(x2 + y2 i) = x1 x2 + (x1 y2 + y1 x2 )i + y1 y2 (−1) E O = (x1 x2 − y1 y2 ) + (x1 y2 + y1 x2 )i. M For real numbers x1 , x2 , y1 , y2 , so we define the number x1 + y1 i as follows. G U A complex number is an ordered pair (x, y) of real numbers with equality and N operations sum and product of two complex numbers (x1 , y1 ), (x2 , y2 ) are defined as follows: Let z1 = x1 + y1 i = (x1 , y1 ) and z2 = x2 + y2 i = (x2 , y2 ). 3  Grade 12 Mathematics Textbook Equality z1 = z2 , if and only if x1 = x2 and y1 = y2. Sum z1 + z2 = (x1 + y1 i) + (x2 + y2 i) = (x1 + x2 ) + (y1 + y2 )i = (x1 + x2 , y1 + y2 ). E M 𝐏𝐫𝐨𝐝𝐮𝐜𝐭 z1 z2 = (x1 + y1 i)(x2 + y2 i) = x1 x2 + x1 y2 i + x2 y1 i − y1 y2 G = (x1 x2 − y1 y2 ) +(x1 y2 + x2 y1 )i U N = (x1 x2 − y1 y2 , x1 y2 + x2 y1 ). Then we have, (x, 0) + (y, 0) = (x + y, 0) and (x, 0)(y, 0) = (xy, 0). E O Real Numbers Complex Numbers M x (x, 0) y G (y, 0) U N x+y (x, 0) + (y, 0) E xy (x, 0)(y, 0) O Table 1.1 M G i = (0, 1) and i2 = (0, 1)(0, 1) U = (0 − 1, 0 + 0) N = (−1, 0) = −1. E We have O x + yi = (x, 0) + (y, 0)(0, 1) M = (x, 0)(0 − 0, y + 0) G = (x, 0) + (0, y) U x + yi = (x, y). N Note that sum and product of complex numbers satisfy commutative, associative and distributive properties. 4  Grade 12 Mathematics Textbook Complex Number A number that has both a real part and an imaginary part. The imaginary part is a multiple of the square root of minus one (i). Some algebraic equations cannot be solved with real numbers. E M Example 3 G Compute (−2, 3)(1, − 2) + (1, 1)(0, 1). U Solution N (−2, 3)(1, − 2) + (1, 1)(0, 1) = (−2 + 3i)(1 − 2i) + (1 + i)i E = −2 + 4i + 3i − 6i2 + i+i2 O = −2 + 8i + 6 − 1 M = 3 + 8i = (3, 8). G U N Exercises 1.2 E 1. Compute: O (a) (2, 0)(3, 5) + (3, − 2)(0, 1) M (b) (2, − 5)(−1, 0) + (1, 0)(5, 1) G U (c) (−3, − 2)(−2, − 3) + (−2, − 3)(−3, − 2) N (d) (1, 0)(0, 1) + (0, 1)(1, 0). E O 2. Compute: M (a) (3 + 2i)(3 − 2i) + (−5 + 7i)(−1 − i) G (b) (−1 + i)(1 − i) + (2 + 3i) U N (c) (1 + i)(1 − i) + (−2 + i)(−2 + i) (d) (3 + 2i) + (7 − i)(−3 + 3i). 5  Grade 12 Mathematics Textbook 1.3 Operations on Complex Numbers Sum and product of complex numbers were defined in Section 1.2. Subtraction of complex numbers is defined as in real numbers as follows: E (x1 + y1 i) − (x2 + y2 i) = (x1 + y1 i) + (−x2 − y2 i). M Therefore sum, product and subtraction of complex numbers can be performed as in the G real numbers, except only that i2 = −1. But the division of complex numbers is a little U different from the division of real numbers. We need some notations to define the division. N Let us denote a complex number by z, so that E z = x + yi = (x, y). O M The conjugate z̅ of a complex number z is defined by z̅ = x − yi = (x, −y). Then we have G U zz̅ = (x + yi)(x − yi) = x 2 − y 2 i2 = x 2 + y 2. N Let z1 = x1 + y1 i and z2 = x2 + y2 i. E z We will calculate z1 , (z2 ≠ 0), as follows: O 2 M z1 x1 + y1 i x2 − y2 i =. z2 x2 + y2 i x2 − y2 i G U x1 x2 − x1 y2 i + x2 y1 i + y1 y2 = x2 2 − y2 2 i2 N (x1 x2 + y1 y2 ) + i(x2 y1 − x1 y2 ) E = x2 2 + y2 2 O M x1 x2 + y1 y2 x2 y1 − x1 y2 = 2 2 +i. x2 + y2 x2 2 + y2 2 G U N 6  Grade 12 Mathematics Textbook Example 4 2+3i Calculate. 3+i Solution E M 2 + 3i 2 + 3i 3 − i =. 3+i 3+i 3−i G 6 − 2i + 9i − 3i2 U = 9 − i2 N 6 + 7i + 3 = 9+1 E O 9 + 7i = M 10 9 7 = + i. G 10 10 U N E Exercises 1.3 O M 1. Let z1 = −2 + 3i, z2 = 5 + 2i. Compute: (a) z1 − 2z2 + 1 (b) 3z2 2 + 2z2 + 1 (c) z1 z̅2 + z2 z̅1 G U 1 1 1 (d) (e) (f) N z1 z2 z1 z2 z ̅̅̅ z z (g) z1 (h) z1 (i) z2 E 2 2 1 O (j) (̅̅̅̅̅ z2 ̅̅̅ z1 z2 2z 1 z ) (k) (l) ̅̅̅ + ̅̅̅. M z 1 z1 ̅̅̅ z2 z z1 2 2. Let z1 = 3 − 2i, z 2 = −1 + 4i. Show that G (c) (̅̅̅̅̅ z ̅̅̅ z U ̅̅̅̅̅̅̅̅̅̅̅̅ (a) (z 1 + z2 ) = z̅1 + z̅2 (b) ̅̅̅̅̅̅ z1 z2 = z̅1 z̅2 1 1 ) = ̅̅̅. z 2 z 2 N 7  Grade 12 Mathematics Textbook 1.4 Trigonometric Form As a complex number z = x + iy is an ordered pair (x, y) of real numbers, we can place z in xy −coordinate plane as usual. If the length of the line segment from O to z is r , we say that the absolute value of z is r and is denoted by |z|. Here angle θ, measured in radians, is an E angle with positive x −axis and the line segment. M Let z = x + iy = (x, y) G U Absolute value of z = |z| = r = √x 2 + y 2 N y x = r cos θ, y = r sin θ, tan θ = x E So, we have z = x + iy O = r cos θ + i r sin θ Figure 1.1 M = r(cos θ + i sin θ) (Trigonometric Form) = (r, θ). G (Polar Form) U N Here angle θ, measured in radians, is an angle with positive x-axis and the line segment. E O Argument of a complex number (𝛉) M G y Im(z) tan 𝜃 = x U y 𝜃 = tan−1 ( ) N x (x, y) E y y Argument of θ = tan−1 ( ), (−π < θ ≤ π) O x θ Re(z) O x The angle θ is called the argument of z, arg z. M G Figure 1.2 U N 8  Grade 12 Mathematics Textbook Principle argument of z The value of θ such that −π < θ ≤ π is called the principal argument of z. st nd 𝜃 is the 1 quadrant 𝜃 is the 2 quadrant y E z = x + iy M G 𝜃 x O U N Figure 1.3 Figure 1.4 E O rd th 𝜃 is the 3 quadrant 𝜃 is the 4 quadrant M y y G U o x x N O θ = −𝜙 E O z = x − yi Figure 1.5 M Figure 1.6 G Example 5 U Find the trigonometric form with −π < θ ≤ π. N (a) z = 1 + √3i (b) z = −1 + i (c) z = −√3 − i (d) z = −1 E Solution O (a) z = 1 + √3i = (1, √3) = (x, y) y M x = 1, y = √3 z = (1, √3) √3 G r = √x 2 + y 2 = √1 + 3 = 2 2 U x 1 √3 cos θ = r = 2, sin θ = 𝜃 N 2 x π O θ = 3 , −π < θ ≤ π. 1 z = r(cos θ + i sin θ) π π = 2(cos 3 + i sin 3 ). 9  Grade 12 Mathematics Textbook (b) z = −1 + i = (−1, 1) = (x, y) y z = (−1, 1) r = √x 2 + y 2 = √1 + 1 = √2 1 √2 x −1 1 cos θ = r = , sin θ = √2 √2 θ E 𝜙 x O M π −1 Basic acute angle = ϕ = 4 G π 3π θ=π−ϕ=π−4 =. 4 U N z = r(cos θ + i sin θ) 3π 3π = √2(cos + i sin ). E 4 4 O M (c) z = −√3 − i = (−√3, − 1) = (x, y) y r = √3 + 1 = √4 = 2 G U N −√3 1 −√3 cos θ = , sin θ = − 2 O 2 x 𝜙 2 θ E π Basic acute angle = ϕ = 6 O −1 z = (−√3, − 1) M π 5π θ = −(π − ϕ) = − (π − 6 ) = −. 6 G z = r(cos θ + i sin θ) U 5π 5π = 2 (cos (− ) + i sin (− )). N 6 6 E O (d) z = −1 = −1 + 0i = (−1, 0) = (x, y) M r = √1 + 0 = √1 = 1 G −1 0 cos θ = = −1, sin θ = 1 = 0 U 1 N θ=π z = 1(cos π + i sin π) = cos π + i sin π. 10  Grade 12 Mathematics Textbook Product in Trigonometric Form Let z1 = r1 (cos θ1 + i sin θ1 ) z2 = r2 (cos θ2 + i sin θ2 ) E M z1 z2 = r1 (cos θ1 + i sin θ1 ). r2 (cos θ2 + i sin θ2 ) G = r1 r2 (cos θ1 cos θ2 + i2 sin θ1 sin θ2 ) + i(sin θ1 cos θ2 + cos θ1 sin θ2 ) U N = r1 r2 (cos θ1 cos θ2 − sin θ1 sin θ2 ) + i(sin θ1 cos θ2 + cos θ1 sin θ2 ) So z1 z2 = r1 r2 (cos(θ1 + θ2 ) + i sin(θ1 + θ2 )). E O M Example 6 Given that z1 = 1 + √3i, z2 = −1 + i find z1 z2 by using trigonometric forms. Check your G answer by direct multiplication. U N Solution z1 = 1 + √3i = (1, √3) = (x1 , y1 ) E y O r1 = √1 + 3 = √4 = 2 z = (1, √3) M √3 1 √3 sin θ1 = , cos θ1 = 2 2 2 G π θ1 = 𝜃1 3 x U O 1 N π π z1 = 2(cos 3 + i sin 3 ) z2 = −1 + i = (−1, 1) = (x2 , y2 ) E O r2 = √1 + 1 = √2, y M z = (−1, 1) 1 −1 1 sin θ2 = , cos θ2 = √2 G √2 √2 U π θ ϕ= 𝜙 x N 4 −1 O π 3π θ2 = π − ϕ = π − 4 = 4 3π 3π z2 = √2(cos + i sin ) 4 4 11  Grade 12 Mathematics Textbook π π 3π 3π z1 z2 = 2 (cos 3 + i sin 3 ) √2 (cos + i sin ) 4 4 π 3π π 3π = 2√2(cos (3 + ) + i sin (3 + )) 4 4 13π 13π E = 2√2 (cos + i sin ) 12 12 M −√6−√2 −√6+√2 = 2√2( + i) (use calculator) 4 4 G U = (−1 − √3) + (1 − √3)i (use calculator) N z1 z2 = (1 + √3i)(−1 + i) = −1 + i − √3i + √3i2 E O = −1 + i − √3i − √3 M = (−√3 − 1) + (−√3 + 1)i. G U N Multiplicative Inverse in Trigonometric Form x y Since z −1 = x2 +y2 − x2 +y2 i for z = x + iy, E O 1 z −1 = x2 +y2 (x − iy) M 1 = r2 r (cos θ − i sin θ ) G U 1 = (cos θ − i sin θ ) N r 1 z −1 = (cos(−θ) + i sin(−θ )). r E O M Division in Trigonometric Form Let z1 = r1 (cos θ1 + i sin θ1 ) and z2 = r2 (cos θ2 + i sin θ2 ). G z1 Then = z1 z2 −1 U z2 N r = r1 (cos(θ1 − θ2 ) + i sin(θ1 − θ2 )). 2 12  Grade 12 Mathematics Textbook Example 7 Given that z = −√3 − i, using trigonometric form of z, find z −1. Check your answer by showing that zz −1 = 1. E M Solution z = −√3 − i = (−√3, − 1) = (x, y) G y U r = √3 + 1 = √4 = 2 N −1 −√3 sin θ = , cos θ = −√3 O x E 2 2 𝜙 O π 𝜃 ϕ= 2 M 6 −1 𝑧 = (−√3, −1) θ = − (π − 6 ) = − π 5π 6 G U N 5π 5π z = 2 (cos (− ) + i sin (− )) 6 6 1 E z −1 = r (cos(−θ) + i sin(−θ)) O 1 5π 5π M = 2 cos + i sin 6 6 1 1 G √3 = 2 (− + i 2) 2 U −√3 1 N = + 4i. 4 √3 1 E zz −1 = (−√3 − i) (− + i) 4 4 O M 3 √3 √3 1 = − i+ i − i2 4 4 4 4 G 3 1 = + = 1. U 4 4 N 13  Grade 12 Mathematics Textbook Example 8 z1 Given that z1 = 1 + √3i, z2 = −1 + i, find by using trigonometric forms. z2 Check your answer by direct calculation. y z = (1, √3) E Solution √3 M 2 z1 = 1 + √3i = (1, √3) = (x1 , y1 ) θ G x r1 = √1 + 3 = √4 = 2 O U 1 √3 1 N sin θ1 = , cos θ1 = 2 2 π θ1 = E 3 O π π z1 = 2(cos 3 + i sin 3 ) M z2 = −1 + i = (−1, 1) = (x2 , y2 ) G z = (−1, 1) y U r2 = √1 + 1 = √2 1 N √2 1 −1 sin θ2 = , cos θ2 = √2 √2 θ 𝜙 E π x ϕ= O O 4 −1 M π 3π θ = (π − ϕ) = π − 4 = 4 G 3π 3π z2 = √2 (cos + i sin ) U 4 4 N z1 r = r1 [cos(θ1 − θ2 ) + i sin(θ1 − θ2 )] z2 2 E 2 π 3π π 3π = [cos (3 − ) + i sin (3 − )] √2 4 4 O M 2 π 3π π 3π = [cos (3 − ) + i sin (3 − )] √2 4 4 G 2 5π 5π = [cos (− 12 ) + i sin (− 12 )] √2 U N −1+√3 1+√3 = − i. (use calculator) 2 2 14  Grade 12 Mathematics Textbook z1 1 + √3i −1 − i = × z2 −1 + i −1 − i −1 − i − √3i − √3i2 = 1 − i2 E −1 − i − √3i + √3 = M 1+1 −1 + √3 1 + √3 = − i. G 2 2 U N Powers of Complex Numbers E The power of a complex number z = r(cos θ + i sin θ) is given by O z n = r n (cos n θ + i sin n θ), n is an integer. M G U Example 9 N Given that z = 1 + √3i, find (a) z10 (b) z −10. E Solution O y z = 1 + √3i = (1, √3) = (x, y) M 1 z = (1, √3) √3 r = √1 + 3 = √4 = 2, sin θ = , cos θ = 2 √3 2 2 G π ∴θ= 3 U π π 𝜃 z = 2 (cos + i sin ). x N 3 3 O 1 E 10π 10π (a) z10 = 210 (cos + i sin ) 3 3 O 1 M √3 = 210 (− 2 − i) , (use calculator) 2 1 √3 G = 1024 (− 2 − i) 2 U = −512 − 512√3i. N 15  Grade 12 Mathematics Textbook −10π −10π (b) z −10 = 2−10 (cos + i sin ) 3 3 1 √3 = 2−10 (− 2 + i) , (Use Calculator) 2 E 1 1 √3 = 1024 (− 2 + i) 2 M 1 √3 = − 2048 + 2048 i. G U N Exercises 1.4 E 1. Find the trigonometric form with −π < θ ≤ π. O (a) z = 1 − √3i (b) z = −√2 + √2i (c) z = −2 − 2i M (d) z = √3 − 1 (e) z = i (f) z = −3i G U 2. Given that z1 = 2 − 2√3i, z2 = −1 − i, find the following complex numbers by using N trigonometric forms. Check your answer by direct calculation. z z (a) z1 z2 (b) z1 −1 (c) z2 −1 (d) z1 (e) z2 E 2 1 O M 3. Given that z = −2√3 − 2i, find (a) z 5 (b) z −5. G 1.5 Roots of Complex Numbers U N An nth root of a complex number z is a complex number w that satisfies the equation 1 E θ+2kπ θ+2kπ w n = z, since z = r(cos θ + i sin θ) and wk = r n (cos + i sin ) n n O M where k = 0, 1, 2,... , n − 1. 1 Let w n = z ⟺ z n = w G U n θ+2kπ θ+2kπ wk = √r (cos + i sin ), k = 0, 1, 2,... , n − 1. N n n n θ θ When k = 0, w0 = √r (cos n + i sin n) 16  Grade 12 Mathematics Textbook n θ+2nπ θ+2nπ n θ θ When k = n, wn = √r (cos + i sin ) = √r (cos n + 2π) + i sin n + 2π) n n So, the root for k = n is the same as the root for k = 0. The same is also true for k > n as the root for k = n+1 is the same as the root for k = 1, and so on. E Therefore, the roots of z, denoted by wk of z are M n θ+2kπ θ+2kπ wk = √r (cos + i sin ), k = 0, 1,... , n − 1. n n G U N Example 10 Find the cube roots of z = −2 − 2i. (OR) Solve z 3 = −2 − 2i. E O Solution M y z = −2 − 2i = (−2, − 2) = (x, y) r = √4 + 4 = √8 = 2√2 G U −2 −1 −2 1 sin θ = 2√2 = , cos θ = 2√2 = − −2 O x N √2 √2 𝜙 π θ ϕ= 4 E −2 𝑧 = (−2, − 2) O π π 3π θ = − (π − 4 ) = − (π − 4 ) = − 4 M 3π 3π z = 2√2 [cos (− ) + i sin (− )] 4 4 G 3π 3π U 3 − +2kπ − +2kπ wk = √2√2 [cos 4 + i sin 4 ], k = 0, 1, 2. 3 3 N Therefore, the cube roots are E −3π⁄4 −3π⁄4 O w0 = √2 (cos ( ) + i sin ( )) 3 3 M −3π −3π = √2(cos ( 12 ) + i sin ( 12 )) G π π U = √2 (cos (− 4 ) + i sin (− 4 ) ) N √2 √2 = √2 ( 2 − i) = 1 − i. 2 17  Grade 12 Mathematics Textbook 3π 3π − +2π − +2π 4 4 w1 = √2 [cos + i sin ] 3 3 5π 5π = √2 (cos ( 12 ) + i sin (12 ) ) √6−√2 √6+√2 = √2 ( + i) E 4 4 M −1+√3 1+√3 = + i. 2 2 G 3π 3π − +4π − +4π 4 4 w2 = √2 [cos + i sin ] U 3 3 N 13π 13π = √2 (cos ( 12 ) + i sin ( 12 ) ) −√6−√2 −√6+√2 = √2 ( + i) E 4 4 O −√3−1 −√3+1 = + i. M 2 2 Example 11 G U N Solve z 6 = 1. (OR) Find the sixth roots of unity. Solution E O M 1 = (1, 0), r = √1 + 0 = 1. G cos θ = 1, sin θ = 0 ⟹∴ θ = 0. U 1 = 1(cos 0 + i sin 0) N = 1(cos(0 + 2kπ) + i sin(0 + 2kπ)) E O = 1(cos 2kπ + i sin 2kπ). M z 6 = 1 = 1(cos 2kπ + i sin2kπ) G U 2kπ 2kπ z = 11⁄6 [cos + i sin ] , k = 0, 1, 2, 3, 4, 5. N 6 6 2kπ 2kπ z = cos + i sin. 6 6 18  Grade 12 Mathematics Textbook When k = 0, z = cos 0 + i sin 0 = 1. π π 1 √3 When k = 1, z = cos 3 + i sin 3 = 2 + i. 2 2π 2π 1 √3 When k = 2, z = cos + i sin = −2 + i. E 3 3 2 M When k = 3, z = cos π + i sin π = −1. G 4π 4π 1 √3 When k = 4, z = cos + i sin = −2 − i. 3 3 2 U N 5π 5π 1 √3 When k = 5, z = cos + i sin =2−i. 3 3 2 E O M Exercise 1.5 1. Find the square roots of the following complex numbers. G U N (𝑎) 1 + √3𝑖 (b) 𝑖 (c) −√3 + 𝑖 E (d) −1 − √3𝑖 (e) −𝑖 (f) √3 − 𝑖 O M 2. Find the cube roots of the following complex numbers. (a) 1 + 𝑖 (b) 𝑖 (c)−1 + 𝑖 G U (d) −1 − 𝑖 (e)−𝑖 (f) 1 − 𝑖 N 3. Solve the following equations. E O (𝑎) 𝑧 4 = −𝑖 (b) 𝑧 4 = −1 (c) 𝑧 4 = −8 − 8√3𝑖 (d) 𝑧 6 = −1 M G U N 19  Grade 12 Mathematics Textbook Chapter 2 MATHEMATICAL INDUCTION In this chapter, we will study a technique of proving a statement, theorem or formula E which is thought to be true, for each and every natural number n. By generalizing this in form M of a principle which we would use to prove any mathematical statement is "Principle of Mathematical Induction". G ▪ What is the statement in Mathematics? U In mathematics, a statement is a sentence that is either true or false but not both. N ▪ Which of the following sentences are statements? E O (1) 3 + 4 = 8. M (2) 2x + 5 = 10. (3) 12 is a multiple of 4. G (4) x2 – y2 = (x + y) (x – y) for all real numbers x and y. U (5) The sum of a and b is greater than 0. N In the above (1), (3) and (4) are statements but others are not. E 2.1 Introduction O One of the techniques to prove mathematical statements discussed in this chapter is the M Principle of Mathematical Induction. This method is a powerful and elegant technique for G proving certain type of mathematical statements which assert that something is true for all U natural numbers or for all natural numbers from some point on. The actual term mathematical N induction was first used by De Morgan, even though the method was used by Fermat, Pascal and others before him. The method is used in many branches of Higher Mathematics. E Before we state the principle of mathematical induction, let us consider an example. O Consider the sum of the first n odd positive integers. That is, M if n = 1, 1 = 12, if n = 2, 1+3 = 4 = 22, G = 32, U if n = 3, 1+3+5 = 9 N if n = 4, 1+3+5+7 = 16 = 42, if n = 5, 1+3+5+7+9 = 25 = 52, if n = 6, 1 + 3 + 5 + 7 + 9 +11 = 36 = 62, 20  Grade 12 Mathematics Textbook From the result above, it looks as if the sum of the first n odd natural numbers is always given by n2. To express this statement symbolically, first observe that the nth odd number is 2n – 1. Then the statement can be expressed as: E 1 + 3 + 5 + 7 + 9 +... + (2n – 1) = n2 ….. (1) M Although from this pattern we might conjecture that the statement (1) is true for any choice of n, can we really be sure that it does not fail for some choice of n? We have seen that the G statement is true for n = 1, 2, 3, 4, 5, 6 by direct calculation. Is the statement true for any natural U number n? Actually, no matter how many cases we check, we can never prove that the statement N is always true because there are infinitely many cases and direct calculation cannot check them all. The method of proof by mathematical induction will, in fact, prove that any mathematical E O statement like (1) is true for all natural numbers n. M 2.2 Principle of Mathematical Induction G For each natural numbers n, let P(n) be a statement depending on n. Suppose that the U following two conditions are satisfied. N 1. The statement is true for n = 1. 2. For any natural number k, if the statement is true for E n = k, then the statement is true for n = k + 1. O Then the statement P(n) is true for all natural numbers n. M Motivation : To understand the principle of mathematical induction, suppose a set of thin G rectangular dominos are placed on one end as shown in Figure (2.1). U N E O M G U N Figure 2.1 21  Grade 12 Mathematics Textbook When the first domino is pushed in the indicated direction, all the dominos will fall. To be absolutely sure that all the dominos will fall, it is sufficient to know that (a) The first domino falls, and (b) In the event that any domino falls its successor necessarily falls. E This phenomenon is called the Domino Effect or chain reaction. M This is the underlying principle of mathematical induction. To apply the principle of mathematical induction, there are two steps: G U The initial step : Prove that the statement is true for n = 1. N The inductive step: Assume that the statement is true for n = k , and use this assumption to prove that the statement is true for n = k + 1. E O Conclusion: The statement has been proved for n = 1. By the inductive step, since the M statement is true for n = 1, it is also true for n = 2. Again, by the inductive step, since the statement is true for n = 2, it is also true for n = 3. And since the statement is true for n = 3, it G is also true for n = 4, and so on. Finally, we conclude that the statement is true for all natural U numbers n. N Remark: In the inductive step we do not prove that the statement is true for n = k. E O We only show that if the statement is true for n = k, then the statement is also true M for n = k + 1. The assumption that the statement is true for n = k, is called the inductive hypothesis. G To see the important of these two steps, we will study the following two examples: U N 1st example : E Consider the following table. O n 1 2 3 4 5 6 7 M n2 + n + 11 13 17 23 31 41 53 67 G U N Can we conclude that n2 + n + 11 is prime for all natural numbers. 22  Grade 12 Mathematics Textbook Let P(n) be the statement n2 + n + 11 is prime, for all natural numbers n. It can be easily seen that P(1) is true. But we must prove that if P(k) is true, then P(k + 1) is true for any k > 1. This inductive step breaks down when k = 9. For, E if k = 9, 92 + 9 + 11 = 101 is prime and M if k = 10, 102 + 10 + 11 = 121 is not prime. G So, we cannot conclude that the statement P(n) is true for all natural numbers n. U N 2nd example : Consider the statement 3n + 2 is a multiple of 3 for all natural numbers n. E Let P(n) be the statement 3n + 2 is a multiple of 3 for all natural numbers n. O We will only prove the inductive step. M Assume that P (k) is true. i.e, 3k + 2 is a multiple of 3. This can be written as the equation G U 3k + 2 = 3m for some natural number m. N We want to show that P (k + 1) is true. For n = k + 1, E 3 (k + 1) + 2 = 3k + 3 + 2 O = 3k + 2 + 3 M = 3m + 3 G = 3(m + 1). U Since m + 1 is a natural number, N 3(k + 1) + 2 is a multiple of 3. ∴ P (k + 1) is true. E O The above proof shows that the inductive step is true. Does it follow that P(n) is true for M all natural numbers n ? No, the statement P(n) is false because 3n + 2 cannot be written as a multiple of 3. G Although the inductive step is true, we cannot conclude that the statement is true. Because we have not established the basis step, the induction fails. In fact, P(1) is false. Since the first U domino does not fall, we cannot even start the chain reaction. N The above two examples show that both steps in mathematical induction are equally important. 23  Grade 12 Mathematics Textbook Example 1 Use the mathematical induction principle to prove that 1 + 3 + 5 + · · · + (2n – 1) = n2, for all natural numbers n. E Solution M Let P(n) denote the statement 1 + 3 + 5 + · · · + (2n – 1) = n2 for all natural numbers n. G For n = 1, L.H.S = 1, U R.H.S = 12 = 1. N ∴ L.H.S = R.H.S E O ∴ P (1) is true. M Assume that P(k) is true. i.e., 1 + 3 + 5 + · · · + (2k – 1) = k2...... (1) G U We will show that P (k + 1) is true. N For n = k + 1, L.H.S = 1 + 3 + 5 +... + (2(k + 1) – 1) E = 1 + 3 + 5 +... + (2k – 1) + (2k + 1) O = k 2 + 2k + 1 (From (1)) M = (k + 1)2 R.H.S = (k + 1)2 G  L.H. S = R.H.S U  P (k + 1) is true. N Hence, by the principle of mathematical induction, E the statement P(n) is true for all natural numbers n. O M Example 2 n(n + 1) Use the mathematical induction principle to prove that 1 + 2 + 3 + +n = , for all G 2 natural numbers n. U N Solution n(n + 1) Let P(n) denote the statement 1 + 2 + 3 + +n= for all natural numbers n. 2 For n = 1, L.H.S = 1, 24  Grade 12 Mathematics Textbook 1(1 + 1) R.H.S = = 1. 2 ∴ L.H.S = R.H.S ∴ P(1) is true. Assume P(k) is true. E k(k+1) i.e., 1 + 2 + 3 + ⋯+ k = …… (1) M 2 We want to show that P(k + 1) is true. G For n = k + 1, L.H.S = 1 + 2 + 3 +... + (k + 1) U N = 1 + 2 + 3 +... + k + (k + 1) k(k+1) = + (k + 1) (From (1)) 2 E k = (k + 1)(2 + 1) O M (k+1)(k+2) = 2 (k+1)(k+1+1) R.H.S = G 2 U (k+1)(k+2) = 2 N ∴ L.H.S = R.H.S ∴ P (k +1) is true E O Hence, by mathematical induction, the statement P(n) is true for all natural numbers n. M G Example 3 U Use the mathematical induction principle to prove that (ab)n = an bn for every natural numbers n. N Solution Let P(n) denote the statement (ab)n = anbn. E For n = 1, L.H.S = (ab)1 = ab O R.H.S = a1 b1 = ab. M ∴ L.H.S = R.H.S G ∴ P (1) is true. U Assume that P(k) is true. N i.e. (ab)k = ak bk..... (1) We want to show that P(k + 1) is true. 25  Grade 12 Mathematics Textbook For n = k + 1, L.H.S = (ab)k+1 = (ab)k (ab) = (ak bk) (ab) (From (1)) = (ak a) (bk b) E = ak+1 bk+1 M R.H.S = ak+1 bk+1 ∴L.H.S = R.H.S G U ∴ P(k + 1) is true. N Hence, by mathematical induction, the statement P(n) is true for every natural numbers n. E Example 4 O (2n−1)3n+1 +3 Prove that 1 ⋅ 3 + 2 ⋅ 32 + 3 ⋅ 33 + ⋯ + n ⋅ 3n = M