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COMPLEX NUMBER P.M./Complex/p.1 Define i 2 = -1 . A complex number z can be expressed in the form z = x + iy where x, y R and x - real part of z (Re z) ; y - imaginary part of z (Im z) : imaginary axis i - purely imaginary unit . The complex number z = x + iy is represented by the point (x, y) on the complex plane / Gaussian plane /Argand × (x , y) diagram. 0 Note : real axis 1. For a real number x (x = x + i 0), it is represented by a point lying on the real axis in complex plane. 2. A complex number z is purely imaginary iff Re z = 0 , and the point representing z lies on the imaginary axis. 3. Let P be the point representing the complex number z = x + iy on the complex plane. y (i) modulus of z : imaginary axis x2 + y2 z = (ii) argument or amplitude of z : x (x, y) z arg z = the angle counted from the positive real axis to the vector OP 0 (anti-clockwise : +ve ; clockwise : -ve) = tan -1 ☼ 4. y x Principle range of is (-π, π]. z = x + iy - standard form = r (cos + i sin ) or r cis = r e i where r = z - polar form - exponential form where ei = cos + i sin - Euler's Formula ex = ( xr x2 x3 r ! = 1 + x + 2 ! + 3 ! + ... r=0 eix = 1 + ix + ∴ (ix ) 2 (ix ) 3 + + ... 2! 3! cos x + i sin x = 1 x2 x4 + cos x = 1 2! 4! x2 x4 + - ... + i 2! 4! - ... and sin x = x x x3 x5 + - ... 3! 5! x3 x5 + - ... . ) 3! 5! x real axis Example P.M./Complex/p.2 -1 = _____________________________________________________ ln (-1) = __________________________________________________ ∴ ln (-8) = __________________________________________________ Express the following complex numbers in the polar form : (a) 2i (b) (1 + i)i (c) 1 + i 2 - i (d) (4 + i)(3 - i) Solution (a) ______________________________________________________________ ______________________________________________________________ (b) ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ (c) ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ (d) ______________________________________________________________ ______________________________________________________________ (e) ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ Five Arithmetic Operations (a) Addition/ Subtraction : z1 = x1 + i y1 and z 2 = x 2 + i y 2 where x1 , y1 , x 2 , y 2 R z1 z2 = (x1 x2 ) + i(y1 y2 ) (b) Multiplication : z1 z2 = (x1 + iy 1)(x2 + iy 2 ) (x1x2 - y1y2 ) + i(x1y2 + y1x2 ) i i z1 z2 = r1e 1 r2e 2 i( ) = r1r2e 1 2 = z1z2 = r1r2 and arg z 1z2 = arg z1 + arg z 2 n and arg z n = n arg z Note : z1n = z1 1 1 z1z2 = z1 z2 for any integer n. (e) i Division : P.M./Complex/p.3 z1 x + iy 1 x2 - iy 2 = 1 z2 x2 + iy 2 x2 - iy 2 = (x1x2 + y1y2 ) + i(x2 y1 - y2 x1 ) x22 + y22 = (x1x2 + y1y2 ) (x y - y2 x1 ) + i 2 1 x22 + y22 x22 + y22 z1 1 = (x1x2 + y1y2 )2 + (x2 y1 - y2 x1 )2 2 2 z2 x2 + y2 = = 1 1 2 z 2 = (x1x2 )2 + (y1y2 )2 + (x2 y1)2 + (y2 x1)2 x2 2 + y2 2 (x 2 + y 2)(x 2 + y 2 ) 2 2 1 1 z1 z2 i r e 1 z arg 1 = arg 1 z i 2 r e 2 2 r i ( - ) = arg 1 e 1 2 = arg z1 - arg z 2 r 2 (c) Conjugation : z = x + iy conjugate of z , z = x - iy Properties : x2 + (-y)2 = z (i) z = (ii) arg z = tan -1 (iii) z = (x - iy) = x + iy = z (iv) z = 0 -y = - arg z x (-y) = 0 x = 0 x = y = 0 ∴ (v) (vi) (vii) z = 0 iff z = 0 y = -y z = z x + iy = x - iy ∴ z is self-conjugate , i.e. z = z iff z is real . z z = (x + iy )(x - iy ) = x2 + y2 = z 2 z z = z z 1 2 1 2 (viii) z z = z z 1 2 1 2 (ix) (x) z z 1 = 1 z z 2 2 z + z = 2 Re z z - z = 2i Im z n n z1 = z1 y = 0 z +z 1 2 Proof : 2 z1 + z2 z1 + z2 Triangular Inequality : = z +z z +z 1 2 1 2 = P.M./Complex/p.4 z1 + z2 z1 + z2 = z z +z z +z z +z z 11 1 2 2 1 2 2 ∴ 2 + 2 Re z z + z 1 2 2 z 1 z 1 = z 1 = z1 2 + 2 z1 z2 + z2 2 2 2 +2 z z + z 1 2 2 2 = ( z z = z z ) 1 2 1 2 2 +2 z z + z 1 2 2 2 = z + z 1 2 2 z1 + z2 z1 + z2 OR imaginary axis z1 + z 2 z2 z1 0 z1 - z 2 Claim : real axis z1 + z 2 Proof __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ Example 1. Let z1 , z2 , ... , zn be arbitrary complex numbers . z1z2 + z1z2 . (a) Prove that z1 z1 + z2 z2 (b) Using (a), or otherwise, show that z1 2 + z2 2 + ... + z n 2 Re z1 z 2 + z 2 z 3 + ... + z n-1 z n + z n z1 _________________________________________________________________ P.M./Complex/p.5 ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ 2. (a) Suppose u, v are 2 non-zero complex numbers such that u + v + 1 = 0. Show that u = v = 1 iff 1 1 + + 1 = 0. u v Hence, or otherwise, show that u = v = 1 iff u2 + v2 + 1 = 0 . (b) Let A, B and C be 3 distinct non-collinear points on the complex plane representing the complex numbers z1 , z2 , z3 respectively. By using the result of (a) to show that △ABC is an equilateral △ iff z12 + z2 2 + z32 = z1z3 + z3z2 + z2 z1 . Solution : (a) ' Only if ' assume u = v = 1 , then _____________________________________________ ____________________________________________________________________ ____________________________________________________________________ 1 1 ' If ' assume + + 1 = 0 u v ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ (u + v + 1)2 = 0 _____________________________________________ 1 1 + + 1 = 0 u v _____________________________________________ _____________________________________________ _____________________________________________ _____________________________________________ (b) P.M./Complex/p.6 imaginary axis z1 = x1 + iy 1 and z 2 = x2 + iy 2 z1 - z2 = (x1 - x2 ) + i(y1 - y2 ) z1 - z2 = C (x1 - x2 )2 + (y1 - y2 )2 B = the distance between the points A representing z1 and z 2 3. 0 real axis ∴ A, B and C form an equilateral △ iff __________________________________________________________________ iff __________________________________________________________________ iff __________________________________________________________________ iff __________________________________________________________________ iff __________________________________________________________________ In the Argand diagram, PQR is an equilateral △ whose circumcentre is at the origin. If P represents the complex number 4 + i, find the complex numbers represented by Q and R. Solution : imaginary axis By definition, ∠POQ = ∠QOR = ∠ROP = __________ 4 + i = 17 e i the complex number represented by Q = ____________________________________________ P(4,1) Q 0 real axis = ____________________________________________ = ____________________________________________ = ____________________________________________ R the complex number represented by R = ____________________________________________ = ____________________________________________ = ____________________________________________ Classwork (99 II Q.5) ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ Geometrical Applications of Complex Numbers 1. z1 = x1 + iy 1 and z 2 = x2 + iy 2 P.M./Complex/p.7 imaginary axis z1 + z2 = (x1 + x2 ) + i(y1 + y2 ) z1 - z2 = z1 + (-z2 ) z2 arg z2 - z1 z1 arg ( z2 - z1) arg z1 - z2 = the angle the vector from z1 to z 2 makes with real axis + ve real axis arg ( z1 - z2 ) = the angle the vector from z2 to z1 makes with + ve real axis Note : (i) (ii) 2 -ve arg ( z1 - z2 ) = +ve if counted clockwisely if counted anti-clockwisely arg ( z1 - z2 ) = arg ( z2 - z1) + = arg (z1 - z2 ) imaginary A(z1) axis = arg (z3 - z2 ) C(z3) ∠ABC = - B(z 2) = arg (z1 - z2 ) - arg (z3 - z2 ) z -z = arg 1 2 z - z 3 2 Note : = Case 1 : = z - z arg 2 4 z -z 1 4 Case 2 : z - z and = arg 2 3 z -z 1 3 imaginary axis z4 z3 z - z z - z arg 2 4 - arg 2 3 = 0 z -z z -z 1 3 1 4 z - z arg 2 4 z -z 1 4 z - z 2 4 z -z 1 4 real axis 0 z -z 1 3 =0 z - z 2 3 z2 z1 0 real axis z -z 1 3 is real and positive z - z 2 3 + (-) = z - z arg 2 4 z -z 1 4 imaginary axis z4 z -z 1 3 =π z - z 2 3 z - z z - z 4 1 3 is real and negative 2 z - z z - z 1 4 2 3 z2 - z3 z1 0 real axis Note : For the distinct points representing z1 , z2 , z3 , z4 to form a cyclic quadrilateral, z - z z - z 2 4 1 3 z - z z - z 1 4 2 3 must be real and non-zero. 3. z is a variable complex number moving on the complex plane in such a way that P.M./Complex/p.8 z + 1 . arg = z - 1 3 What is the locus of z ? Solution : ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ 4. Circle with centre representing z0 and radius r : _________________________________________________ 5. Perpendicular bisector of two fixed points representing z1 and z 2 : _________________________________________________ 6. Ellipse with two fixed points representing z1 and z 2 as two foci : _________________________________________________ _________________________________________________ 7. Hyperbola with two fixed points representing z1 and z 2 as two foci : _________________________________________________ Example 1. Find the locus of the variable point P which represents the complex number z such that 2 z - 2 = z - 6i . Solution : ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ 2. Find the locus of the point P representing the complex number z such that z = a + b(1 + it ) 1 - it P.M./Complex/p.9 where a, b are real constants and t is a real variable. Solution : ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ Classwork 3. (01) ___________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ Transformation Suppose two complex variables w and z are connected by a functional relation. Then if the point P representing z describes a curve in an Argand diagram (called the z-plane), the point representing w will describe a related curve in another Argand diagram (called the w-plane). Such a relation from a figure on the z-plane to a figure on the w-plane is called a transformation. Example 1. Let z be a complex variable with z = 1. If w = 3z + 1 , find the locus of the points representing z w. ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ _____________________________________________________________ P.M./Complex/p.10 ________________________________________________________________________ ________________________________________________________________________ 2. ________________________________________________________________________ (90 I Q.10) 1 Let f : C\0 C be defined by f (z) = z + . z Let z = r(cos + i sin ) and f(z) = u + iv where r > 0 and , u , v R, (a) express u and v in terms of r and . (b) Find and sketch the image of each of the following circles under f : (i) z = 1 ; (ii) z = a where 0 < a < 1 . (a) ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ (b) (i) ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ (ii) ________________________________________________________ ________________________________________________________ ________________________________________________________ ________________________________________________________ ________________________________________________________ ________________________________________________________ 3. (89 I Q.12) The function f : C\-1 C\-i is defined by f(z) = (b) i(1 - z) . 1+z Find and sketch the image, under f, of each of the following : (i) the upper half of the imaginary axis (including the origin) ; (ii) the positive real axis . Solution : (i) ___________________________________________________________ ___________________________________________________________ ___________________________________________________________ ___________________________________________________________ ___________________________________________________________ ___________________________________________________________ P.M./Complex/p.11 ___________________________________________________________ ___________________________________________________________ ___________________________________________________________ (ii) ___________________________________________________________ ___________________________________________________________ ___________________________________________________________ ___________________________________________________________ ___________________________________________________________ ___________________________________________________________ W.34 Q.6, 7 H.W. : W.34 Q.1 – 5 De Moivre's Theorem If z = r(cos + i sin ) where r = z , for any rational number n, then zn = r n(cos + i sin )n = r n(cos n + i sin n ) . Proof : Case 1 : Let n be a positive integer. When n = 1, the statement is obviously true. Assume zk = r k (cos k + i sin k ) . ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ Case 2 : Let n be a positive rational number, i.e. n = p and q. Let z p q p q = r (cos + i sin ) p for some positive relatively prime integers q where - < . ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ Case 3 : Let n be a negative rational number, then (-n) is a positive rational number. P.M./Complex/p.12 ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ∴ The De Moivre's Theorem is true for all rational numbers n. Classwork 1. 5 3 + i sin 3 cos - i sin Prove that = cos 13 - i sin 13 . 5 5 cos 5 + i sin 5 cos 2 - i sin 2 cos 3 Solution : ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ 2. 1 + sin + i cos Show that 1 + sin - i cos n = cos n - + i sin n - . 2 2 Solution : ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ Remarks : zn = 1. ∴ 2. z n cos n zn = z + i sin n = z n cos n + i sin n n if 0 z < 1 , then lim n zn = lim n z n =0 P.M./Complex/p.13 lim n zn = 0 imaginary axis x z x z2 where z n is the distance of the point representing z n x from the origin . n x z 0 x z3 real axis Application of De Moivre's Theorem to Trigonometry I. 1 - zn 1 + z + z2 + ... + z n -1 = 1- z if z < 1 , then zr = r=0 1 1 - z (i.e. zr converges if z < 1 ) r=0 Example 1 To find r=0 2 r cos r . Consider ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ II. Express cosine or sine of multiple angles in terms of cos , sin , i.e. cos m = f (cos, sin ) sin m = f (cos, sin ) . Example (a) Prove that cos 5 = 16 cos5 - 20 cos3 + 5 cos . 2 4 6 8 1 (b) Hence, deduce that cos cos cos cos = 5 5 5 5 16 Solution : (a) ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ _____________________________________________________________ P.M/Complex/p.14 ________________________________________________________________________ ________________________________________________________________________ (b) ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ _____________________________________________ _____________________________________________ _____________________________________________ 4 5 2 5 unit circle 0 _____________________________________________ _____________________________________________ 6 5 _____________________________________________ 8 5 ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ Classwork (a) Prove that cos 6 = 32 cos6 - 48 cos4 + 18 cos2 - 1. (b) Hence, show that the roots of the equation 64x3 - 96x2 + 36x - 3 = 0 are 5 7 5 7 and deduce that sec2 cos2 , cos2 , cos2 + sec2 + sec2 = 12 . 18 18 18 18 18 18 (a) ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ (b) ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ III. Express a power of sin or cos in terms of sines or cosines of multiple angles. P.M/Complex/p.15 1. De Moivre's Theorem 2. Binomial Theorem 3. Let z = cos + i sin z n = cos n + i sin n cos n = z n + z-n 2 z = z = 1 n ; z -n = z and 1 z = cos n - i sin n sin n = z n - z-n 2i Example Show that 32 cos6 = cos 6 + 6 cos 4 + 15 cos 2 + 10 . Solution : 64 cos6 = _______________________________________________________________ = _______________________________________________________________ = _______________________________________________________________ = _______________________________________________________________ = _______________________________________________________________ Advantage : easier to evaluate e.g. 32 cos 6 d cosn x dx = _________________________________________________________ = _________________________________________________________ Classwork (a) Show that 16 sin 5 = sin 5 - 5 sin 3 + 10 sin . (b) Hence, solve the equations (i) 16 sin 5 = sin 5 ; 1 (ii) . cos 5 + 5 cos 3 + 10 cos = 2 Solution : (a) ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ (b)(i)________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ (ii) _______________________________________________________________ P.M/Complex/p.16 ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ W.35 Ex.A Q.4 H.W. Ex.A Q.2 , 3 , 5 The n th Roots of Unity zn - 1 = 0 - a polynomial of degree n , so there should be n roots 1 = cos 2k + i sin 2k where k is any integer 2k 2k Consider z = cos + i sin . k n n Raising both sides to power n, we get 2k 2k z = cos + i sin k n n n n = cos 2k + i sin 2k = 1 z k is a solution of z n - 1 = 0 . Since k can be any integer, there seems to be an infinite number of roots (more than n roots) for the give equation. However, if n is used to divide k, then k = nq + r for some integer q and r = 0, 1, 2, ... , n-1 . 2k 2k 2(nq + r) 2(nq + r) cos + i sin + i sin = cos n n n n 2r 2r = cos 2q + + i sin 2q + n n 2r 2r = cos + i sin n n As r = 0, 1, 2, ... , n-1 , it shows that there are only n roots (n values of zk ) for the given equation. 2k 2k + i sin z = cos k n n or i e 2 k n z 3 z 2 n 2 2 n 2 n for k = 0, 1, 2, ... , n-1 . n th roots of a complex number : z 2 n 1 z 0 z n-1 P.M./Complex/p.17 The n th roots of a complex number w are the n values of z which satisfy the equation zn - w = 0 (*) Let w = r (cos + i sin ) . 1 + i sin . Consider z' = r n cos n n (z' ) n = r cos + i sin = w ∴ z' is a solution of the equation (*) . Let z0 , z1 , ... , zn-1 be the n th roots of 1. The roots of z n - w = 0 are z' z0 , z' z1 , ... , z' zn-1 . ( z' z k ∴ n = (z' ) n ( z k ) n = w 1 = w ) The n th roots of w are 2 k 1 i i n n r e e n for k = 0, 1, 2, ... , n-1 2 k 1 i n = r e n 1 = rn 2k + 2k + + i sin cos n n Equation reducible to the form z n - w = 0 : (z + 1)11 - (z - 1)11 = 0 e.g. Solution : ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ (z + 1)11 - (z - 1)11 = _______________________________________________________ = _______________________________________________________ = _______________________________________________________ = _______________________________________________________ Example : Factorize z2n + z2n-1 + ... + z + 1 into real quadratic factors. P.M/Complex/p.18 Solution : z2n+1 - 1 z2n + z2n-1 + ... + z + 1 = z - 1 Suppose z2n + z2n-1 + ... + z + 1 = 0 , z2n+1 - 1 = 0 and z 1 z = ___________________________________________________________ z2n + z2n-1 + ... + z + 1 ∴ = _______________________________________________________________ = _______________________________________________________________ = _______________________________________________________________ Example : Let n be a positive integer. By solving the equation x2n + 1 = 0 , show that x n + x -n = n (2k - 1) x+ x -1 - 2 cos 2n . k=1 n (2k - 1) Hence, deduce that 22n-1 sin 2 = 1 and 4n k=1 cos n = n (2k - 1) 1 - sin 2 2 csc 2 4n . k=1 Solution : ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ∴ x2n + 1 = ______________________________________________________________ = ______________________________________________________________ = ______________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ (b) ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ______________________________________________________________ P.M/Complex/p.19 ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ Note : n th roots of unity : 2k 2k z = cos + i sin k n n zn = 1 i Let w = e 2 n or i e 2 k n for k = 0, 1, 2, ... , n-1 zk = w k . The n th roots of unity are 1 , w, w 2 , ... , w n-1. For z n - 1 = 0 , sum of roots = - coefficient of z n -1 coefficient of z n 1 + w + w 2 + ... + w n-1 = 0 . Example : x3 - 1 = 0 Let w = e i 2 3 . 1 + w + w2 = 0 Example : n C0 Solution : i + n C3x3 + n C6 x6 + ... = ? 2 3 1 + w + w2 = 0 and w 3 = 1 Let w = e (1 + x) n = n C0 + n C1x + n C2 x2 + n C3x3 + n C4 x4 + ... _______________________________________________________________________ _______________________________________________________________________ _______________________________________________________________________ Therefore, nC0 + n C3 + n C6 + ... = __________________________________________________________________ = __________________________________________________________________ = __________________________________________________________________ = __________________________________________________________________ = __________________________________________________________________ Classwork P.M/Complex/p.20 nC2 + n C5 + n C8 + ... = ? (1 + x) n = n C0 + n C1x + n C2 x2 + n C3x3 + n C4 x4 + ... _______________________________________________________________________ _______________________________________________________________________ _______________________________________________________________________ Therefore, n C 2 + n C 5 + n C 8 + ... = __________________________________________________________________ = __________________________________________________________________ = __________________________________________________________________ = __________________________________________________________________ = __________________________________________________________________ W.35 Ex.B Q.2, 4, 5 H.W. : Ex.B Q.1, 7, 9 Ex.C Q.2 Primitive Root An n th root of unity z is said to be primitive if there doesn't exist a positive integer m less than n for which z m = 1 . z9 = 1 e.g. Consider z3 = 2 i For z3 = e 3 ∴ 2 i 3 zk = e , i e 2k 9 and z 5 = z33 = ei2 for k = 0, 1, 2, ... , 8 10 i 9 . e = 1. There exists a positive integer n less than 9 such that z Let n be a positive integer such that z5n = 1 10 10 cos n + i sin n=1 9 9 10 cos n =1 9 10 n = 2k 9 5n = k 9 ∴ 9 divides n. ∴ 9<n and e n 3 10 i n 9 =1 . =1 10 sin n = 0 9 for some integer k z5 is a primitive root of unity. Theorem : e i 2k n is a primitive n th root of 1 if k, n are relatively prime to each other . P.M./Complex/p.21 2k i Assume 1 = e n cos 2km = 1 n m = and 2km is an even integer n e i 2km 2km n = cos n sin 2km . n + i sin 2km = 0 n km is also an integer n If k, n are relatively prime to each other, then n divides m. n m Therefore, if k and n are relatively prime to each other, then W.35 Ex.C Q.1,3 W.36 Q.1 W.36.5 Q.4 , 9 - 11 , 15 H.W. : W.36 Q.2, 3 W.36.5 Q. 5 ,6 , 7 , 12 , 13 , 14 ,17 END e i 2k n is a primitive n th root of 1 .