Mathematics

1. Mathematics

2. Session Complex Numbers

3. Session Objectives

4. Session Objective • Polar form of a complex number • Euler form of a complex number • Representation of z1+z2, z1-z2 • Representation of z1.z2, z1/z2 • De-Moivre theorem • Cube roots of unity with properties • Nth root of unity with properties

5. z(x,y) Y y  O X x Representation of complex number in Polar or Trigonometric form z = x + iy This you have learnt in the first session

6. z(x,y) Y r y =rsin  O X x = rcos Representation of complex number in Polar or Trigonometric form z = r (cos  + i sin ) Examples: 1 = cos0 + isin0 -1 = cos  + i sin i = cos /2 + i sin /2 -i = cos (-/2) + i sin (-/2)

7. Eulers form of a complex number z = x + iy z = r (cos  + i sin ) Express 1 – i in polar form, and then in euler form Examples:

8. Properties of eulers form ei

9. The value of ii is ____ • 2 b) e-/2 • c)  d) 2 Illustrative Problem Solution: i = cos(/2) + i sin(/2) = ei/2 ii = (ei/2)i = e-/2

10. Illustrative Problem Find the value of loge(-1). Solution: -1 = cos  + i sin  = ei loge(-1) = logeei = i General value: i(2n+1), nZ As cos(2n+1) + isin(2n+1) = -1

11. If z and w are two non zero complex numbers such that |zw| = 1, and Arg(z) – Arg(w) = /2, then is equal to a) i b) –i c) 1 d) –1 Illustrative Problem

12. Solution

13. z(x1+x2,y1+y2) z2(x2,y2) z1(x1,y1) B O A Representation of z1+z2 z1 = x1 + iy1, z2 = x2 + iy2 z = z1 + z2 = x1 + x2 + i(y1 + y2) Oz1 + z1z  Oz ie |z1| + |z2|  |z1 + z2|

14. z2(x2,y2) z1(x1,y1) O z(x1-x2,y1-y2) -z2(-x2,-y2) Representation of z1-z2 z1 = x1 + iy1, z2 = x2 + iy2 z = z1 - z2 = x1 - x2 + i(y1 - y2) Oz + z1z  Oz1 ie |z1-z2| + |z2|  |z1| also |z1-z2| + |z1|  |z2|  |z1-z2|  ||z1| - |z2||

15. r1r2ei(1+ 2) Y 1+2 r2ei2 r1ei1 2 1 O x Representation of z1.z2 z1 = r1ei1, z2 = r2ei2 z = z1.z2 = r1r2ei(1+ 2)

16. Y r1ei(1+ ) r1ei1  1 O x Representation of z1.ei and z1.e-i z1 = r1ei1 z = z1. ei = r1ei(1+ ) What about z1e-i  r1ei(1- )

17. 2 r1ei1 1 r2ei2 r1/r2ei(1- 2) 1- 2 Representation of z1/z2 z1 = r1ei1, z2 = r2ei2 z = z1/z2 = r1/r2ei(1- 2)

18. If z1 and z2 be two roots of the equation z2 + az + b = 0, z being complex. Further, assume that the origin, z1 and z2 form an equilateral triangle, then • a2 = 3b b) a2 = 4b • c) a2 = b d) a2 = 2b Illustrative Problem

19. z2 = z1ei /3 z1 /3 Solution Hence a2 = 3b

20. De Moivre’s Theorem 1) n  Z,

21. De Moivre’s Theorem 2) n  Q, cos n + i sin n is one of the values of (cos  + i sin )n Particular case

22. Illustrative Problem Solution:

23. Illustrative Problem Solution:

24. Cube roots of unity Find using (cos0 + isin0)1/3

25.  O 1 2 Properties of cube roots of unity 1,, 2 are the vertices of equilateral triangle and lie on unit circle |z| = 1 Why so?

26. If  is a complex number such that 2++1 = 0, then 31 is • 1 b) 0 • c) 2 d)  Illustrative Problem

27. Solution

28. Nth roots of unity

29. 3 2 4/n  1 2/n n-1 Properties of Nth roots of unity c) Roots are in G.P d) Roots are the vertices of n sided regular polygon lying on unit circle |z| = 1

30. i -1 1 -i Illustrative Problem Find fourth roots of unity. Solution:

31. Illustrative Problem Solution:

32. Class Exercise

33. Express each of the following complex numbers in polar form and hence in eulers form. (a) (b) –3 i Class Exercise - 1 Solution 1. (a)

34. Solution Cont. b) –3i

35. If z1 and z2 are non-zero complex numbers such that |z1 + z2| = |z1| + |z2| then arg(z1) – arg (z2) is equal to (a) –p (b) (c) 0 (d) Class Exercise - 2 Solution (triangle inequality) z1 and z2 are in the same line z1 and z2 have same argument or their difference is multiple of 2 arg (z1) – arg (z2) = 0 or 2n in general

36. Area of area of square OPQR Class Exercise - 3 Find the area of the triangle on the argand diagram formed by the complex numbers z, iz and z + iz. Solution We have to find the area ofPQR. Note that OPQR is a square as OP = |z| = |iz| = OR and all angles are 90°

37. If where x and y are real, then the ordered pair (x, y) is given by ___. Class Exercise - 4

38. Solution = 325 (x + iy)

39. If then prove that Class Exercise - 5 Solution x + y + z = = 0 + i0 = 0 x3 + y3 + z3 – 3xyz = (x + y + z) (x2 + y2 + z2 – xy – yz – zx) = 0 x3 + y3 + z3 = 3xyz

40. Solution Cont. x3 + y3 + z3 = 3xyz

41. Class Exercise - 6 If and then is equal to ___. d) 0 Solution Take any one of the values say

42. Similarly Solution Cont.

43. The value of the expression wherew is an imaginary cube root of unity is ___. Class Exercise - 7 Solution

44. If are the cube roots of p, p < 0, then for any x, y, z, is equal to (a) 1 (b)  (c) 2 (d) None of these Class Exercise - 8 Solution

45. The value of is • –1 (b) 0 • (c) i (d) –i ...(i) k = 0, 1, …, 6 Class Exercise - 9 Solution: roots of x7 – 1 = 0 are

46. ...(ii) Solution Cont. From (i) and (ii), we get

47. If 1, are the roots of the equation xn – 1 = 0, then the argument of is (a) (b) (c) (d) Class Exercise - 10 Solution: As nth root of unity are the vertices of n sided regular polygon with each side making an angle of 2/n at the centre, 2 makes an angle of 4/n with x axis and hence, arg(2) = 4/n

48. Thank you