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Complex Numbers

Complex Numbers. XII – STANDARD MATHEMATICS. PREPARED BY: R.RAJENDRAN. M.A., M . Sc., M. Ed., K.C.SANKARALINGA NADAR HR. SEC. SCHOOL, CHENNAI-21. If n is a positive integer, prove that. If a = cos 2  + isin 2, b = cos2 + isin 2, c = cos 2 + isin 2. Prove that (i).

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Complex Numbers

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  1. Complex Numbers XII – STANDARD MATHEMATICS PREPARED BY: R.RAJENDRAN. M.A., M. Sc., M. Ed., K.C.SANKARALINGA NADAR HR. SEC. SCHOOL, CHENNAI-21

  2. If n is a positive integer, prove that

  3. If a = cos 2 + isin 2, b = cos2 + isin 2, c = cos 2 + isin 2. Prove that (i) a = cos 2 + isin 2 b = cos 2 + isin 2 c = cos 2 + isin 2 abc = (cos 2 + isin 2 )(cos 2 + isin 2) (cos 2 + isin 2) = cos (2 + 2 + 2) + isin (2 + 2 + 2)

  4. = cos ( +  + ) + isin ( +  + ) ……..(1) = cos ( +  + ) – i sin ( +  + ) …….(2) adding (1) and (2) we get

  5. If a = cos 2 + isin 2, b = cos2 + isin 2, c = cos 2 + isin 2. Prove that (ii) = (cos 2 + isin 2 )(cos 2 + isin 2)(cos 2 + isin 2)– 1 = (cos 2 + isin 2 )(cos 2 + isin 2)(cos 2 – isin 2) = (cos 2 + isin 2 )(cos 2 + isin 2)(cos (–2) + isin (–2)) = cos(2 + 2 – 2) + isin(2 + 2 – 2) = cos 2( +  – ) + isin 2( +  – )………….(1)

  6. = [cos 2( +  – ) + i sin 2( +  – )]-1 = cos 2( +  – ) – i sin 2( +  – )………(2) Adding (1) and (2), we get,

  7. Simplify: (cos + isin )0 = 1

  8. Simplify:

  9. Prove that:

  10. If cos  + cos  + cos  = 0 = sin  + sin  + sin  : prove that(i) cos 3 + cos 3 + cos 3 = 3cos( +  + )(ii) sin 3 + sin 3 + sin 3 = 3sin( +  + ) Given: cos  + cos  + cos  = 0 and sin  + sin  + sin  = 0 (cos  + cos  + cos ) + i(sin  + sin  + sin ) = 0 (cos  + isin )(cos  + isin )(cos  + isin ) = 0 Let a = cos  + isin  b = cos  + isin  c = cos  + isin 

  11. If a + b + c = 0, then a3 + b3 + c3 = 3abc (cos  + isin )3 + (cos  + isin )3 + (cos  + isin )3 = 3(cos  + isin )(cos  + isin )(cos  + isin ) (cos 3 + isin 3) + (cos 3 + isin 3) + (cos 3 + isin 3) = 3[cos ( +  + ) + isin ( +  + )] (cos 3 + cos 3 + cos 3) + i(sin 3 + sin 3 + sin 3) = 3cos ( +  + ) + 3isin ( +  + ) Equating real and imaginary parts cos 3 + cos 3 + cos 3 = 3cos ( +  + ) sin 3 + sin 3 + sin 3 = 3sin ( +  + )

  12. If n is a positive integer, prove that(i) (1 + i)n + (1 – i)n = Let 1 + i = r(cos  + isin )

  13. Substituting i = - i Adding (1) and (2), we get

  14. If n is a positive integer, prove that(ii) (3 + i)n + (3 – i)n = Let 3 + i = r(cos  + isin )

  15. Substituting i = –i Adding (1) and (2), we get

  16. If n is a positive integer, prove that(1 + cos + isin )n + (1 + cos  - isin )n (1 + cos + isin )n + (1 + cos  - isin )n

  17. If  and  are the roots of x2 – 2x + 4 = 0 prove that n – n = i2n+1 sin (n/3) and deduce 9 – 9 x2 – 2x + 4 = 0 a = 1, b = –2, c = 4 The two roots are

  18. Let 1 + i 3 = r(cos  + isin )

  19. Substituting i = –i we get Subtracting (2) from (1) we get

  20. If n = 9 9 – 9

  21. If x + 1/x = 2cos prove that (i)xn + 1/xn = 2cos n (ii) xn – 1/xn = 2isin n

  22. If x + 1/x = 2cos, y + 1/y = 2cos, prove that one of the values of

  23. similarly

  24. Adding (1) and (2)

  25. Solve: x8 + x5 – x3 – 1 = 0 x8 + x5 – x3 – 1 = 0 x5(x3 + 1) – (x3 + 1) = 0 (x3 + 1)(x5 – 1) = 0 x3 + 1 = 0, x5 – 1 = 0 Case (1) x3 + 1 = 0 x3 = – 1 = cos  + isin  = cos(2k + ) + isin(2k + ) x = {cos(2k + ) + isin(2k + )}1/3

  26. The three values are Case (2) x5 – 1 = 0 x5 = 1 = cos 0 + isin 0 = cos(2k) + isin(2k) x = {cos(2k) + isin(2k)}1/5

  27. The five values are

  28. Solve: x7 + x4 + x3 + 1 = 0 x7 + x4 + x3 + 1 = 0 x4(x3 + 1) + (x3 + 1) = 0 (x3 + 1)(x4 + 1) = 0 x3 + 1 = 0, x4 + 1 = 0 Case (1) x3 + 1 = 0 x3 = – 1 = cos  + isin  = cos(2k + ) + isin(2k + ) x = {cos(2k + ) + isin(2k + )}1/3

  29. The three values are Case (2) x4 + 1 = 0 x4 = – 1 = cos  + isin  = cos(2k + ) + isin(2k + ) x = {cos(2k + ) + isin(2k + )}1/4

  30. The four values are

  31. Find all the values of (1 + i)1/4 Let 1 + i = r(cos  + isin )

  32. For k = 0,1,2,3,4 we get all the values

  33. Find all the values of And hence prove that the product of the values is 1 Since sin is – ve and cos is +ve  lies in the fourth quadrant

  34. For k = 0,1,2,3, we get all the values

  35. Their product = = 1

  36. If  and  are the roots of x2 – 2px + (p2 + q2) = 0 and tan = q/ y+p prove that x2 – 2px + (p2 + q2) = 0 a = 1, b = –2p, c = p2 + q2 The two roots are

  37. P represents the variable complex number z, find the locus of P if Let z = x + iy be the variable complex number

  38. x(x + 1) + y(y + 1) = x2 + (y + 1)2 x2 + x + y2 + y = x2 + y2 + 2y + 1 x + y = 2y + 1 x – 2y + y – 1 = 0 x – y – 1 = 0 The locus of P is x – y – 1 = 0

  39. Find the square root of – 8 – 6i. Let square root of – 8 – 6i be x + iy – 8 – 6i = (x + iy)2 = x2 + 2ixy + i2y2 = x2 + 2ixy – y2 – 8 – 6i = x2 – y2 + 2ixy Equating the real and imaginary parts x2 – y2 = – 8 ……..(1) 2xy = –6 ……..(2) xy = –3 y = –3/x Sub y = –3/x in (1)

  40. x4 – 9 = – 8x x4 + 8x2 – 9 = 0 x4 + 9x2 – x2 – 9 = 0 x2(x2 + 9) – 1(x2 + 9) = 0 (x2 + 9)(x2 – 1) = 0 x2 = –9, 1 x = ±3i, ±1 Since is real x = ±1 If x = 1, then y = –3 If x = –1, then y = 3 The square root = 1 – 3i and –1 + 3i

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