1. Trigonometric Form of a Complex Number 8.5

2. 2  Plot complex numbers in the complex plane and find absolute values of complex numbers.  Write the trigonometric forms of complex numbers.  Multiply and divide complex numbers written in trigonometric form. Objectives

3. 3 The Complex Plane

4. 4 Just as real numbers can be represented by points on the real number line, you can represent a complex number z = a + bi as the point (a, b) in a coordinate plane (the complex plane).

5. 5 The Complex Plane The horizontal axis is called the real axis and the vertical axis is called the imaginary axis, as shown below.

6. 6 The Complex Plane The absolute value of the complex number a + bi is defined as the distance between the origin (0, 0) and the point (a, b). When the complex number a + bi is a real number (that is, if b = 0), this definition agrees with that given for the absolute value of a real number | a + 0i | = = | a |.

7. 7 Plot z = –2 + 5i and find its absolute value. Solution: The number is plotted in Figure 4.2. It has an absolute value of Example 1 – Finding the Absolute Value of a Complex Number Figure 4.2

8. 8 Trigonometric Form of a Complex Number

9. 9 You have learned how to add, subtract, multiply, and divide complex numbers. To work effectively with powers and roots of complex numbers, it is helpful to write complex numbers in trigonometric form. In Figure 4.3, consider the nonzero complex number a + bi. Figure 4.3

10. 10 By letting  be the angle from the positive real axis (measured counterclockwise) to the line segment connecting the origin and the point (a, b), you can write a = r cos  andb = r sin  where. Consequently, you have a + bi = (r cos  ) + (r sin  )i from which you can obtain the trigonometric form of a complex number. Trigonometric Form of a Complex Number

11. 11 The trigonometric form of a complex number is also called the polar form. Because there are infinitely many choices for , the trigonometric form of a complex number is not unique. Normally,  is restricted to the interval 0   < 2 , although on occasion it is convenient to use  < 0. Trigonometric Form of a Complex Number

12. 12 Example 2 – Trigonometric Form of a Complex Number Write the complex number z = –2 – 2i in trigonometric form. Solution: The absolute value of z is and the argument angle  is determined from

13. 13 Example 2 – Solution Because z = –2 – 2 i lies in Quadrant III, as shown in Figure 4.4, Figure 4.4 cont’d

14. 14 Example 2 – Solution So, the trigonometric form is cont’d

15. 15 Multiplication and Division of Complex Numbers

16. 16 Multiplication and Division of Complex Numbers The trigonometric form adapts nicely to multiplication and division of complex numbers. Suppose you are given two complex numbers z 1 = r 1 (cos  1 + i sin  1 ) and z 2 = r 2 (cos  2 + i sin  2 ). The product of z 1 and z 2 is z 1 z 2 = r 1 r 2 (cos  1 + i sin  1 )(cos  2 + i sin  2 ) = r 1 r 2 [(cos  1 cos  2 – sin  1 sin  2 ) + i(sin  1 cos  2 + cos  1 sin  2 )].

17. 17 Using the sum and difference formulas for cosine and sine, you can rewrite this equation as z 1 z 2 = r 1 r 2 [cos(  1 +  2 ) + i sin(  1 +  2 )]. This establishes the first part of the following rule. Multiplication and Division of Complex Numbers

18. 18 Note that this rule says that to multiply two complex numbers you multiply moduli and add arguments, whereas to divide two complex numbers you divide moduli and subtract arguments. Multiplication and Division of Complex Numbers

19. 19 Example 6 – Multiplying Complex Numbers Find the product z 1 z 2 of the complex numbers. z 1 = 2(cos 120  + i sin 120  ) z 2 = 8(cos 330  + i sin 330  ) Solution: z 1 z 2 = 2(cos 120  + i sin 120  )  8(cos 330  + i sin 330  ) = 16[cos(120  + 330  ) + i sin(120  + 330  )] = 16(cos 450  + i sin 450  ) Multiply moduli and add arguments.

20. 20 Example 6 – Solution = 16(cos 90  + i sin 90  ) = 16[0 + i(1)] = 16i 450  and 90  are coterminal. cont’d

21. 21 Example 7 – Dividing Complex Numbers Find the quotient z 1  z 2 of the complex numbers. Solution: Divide moduli and subtract arguments.

22. 22 Example 7 – Solution cont’d

23. 23 Multiplication and Division of Complex Numbers

24. 24 Example – Multiplying Complex Numbers in Trigonometric Form Find the product z 1 z 2 of the complex numbers. Solution:

25. 25 Example – Solution = 6(cos  + i sin  ) = 6[–1 + i (0)] = –6 The numbers z 1,z 2 and z 1 z 2 are plotted in Figure 6.52. cont’d Figure 6.52

26. 26 Pages 640 - 641 1 – 39 odd, 47 – 65 odd Homework

27. 27 Powers of Complex Numbers

28. 28 Powers of Complex Numbers The trigonometric form of a complex number is used to raise a complex number to a power. To accomplish this, consider repeated use of the multiplication rule. z = r (cos  + i sin  ) z 2 = r (cos  + i sin  ) r (cos  + i sin  ) = r 2 (cos 2  + i sin 2  ) z 3 = r 2 (cos 2  + i sin 2  ) r (cos  + i sin  ) = r 3 (cos 3  + i sin 3  )

29. 29 Powers of Complex Numbers z 4 = r 4 (cos 4  + i sin 4  ) z 5 = r 5 (cos 5  + i sin 5  ). This pattern leads to DeMoivre’s Theorem, which is named after the French mathematician Abraham DeMoivre (1667–1754).

30. 30 Who was De Moivre? A brilliant French mathematician who was persecuted in France because of his religious beliefs. De Moivre moved to England where he tutored mathematics privately and became friends with Sir Issac Newton. De Moivre made a breakthrough in the field of probability (writing the Doctrine of Chance), but more importantly he moved trigonometry into the field of analysis through complex numbers with De Moivre’s theorem.

31. 31 Powers of Complex Numbers

32. 32 Powers of Complex Numbers This is horrible in rectangular form. The best way to expand one of these is using Pascal’s triangle and binomial expansion. You’d need to use an i-chart to simplify. It’s much nicer in trig form. You just raise the r to the power and multiply theta by the exponent.

33. 33 Example – Finding a Power of a Complex Number Use DeMoivre’s Theorem to find Solution: First convert the complex number to trigonometric form using previous formulas for r and theta: r = = 2 and  = arctan

34. 34 Example – Solution So, the trigonometric form is Then, by DeMoivre’s Theorem, you have (1 + i) 12 cont’d

35. 35 Example – Solution = 4096(cos 4  + i sin 4  ) = 4096(1 + 0) = 4096. cont’d

36. 36 Roots of Complex Numbers There will be as many answers as the index of the root you are looking for Square root = 2 answers Cube root = 3 answers, etc. Answers will be spaced symmetrically around the circle You divide a full circle by the number of answers to find out how far apart they are: Cube root: 360/3 = 120, roots occur spaced 120 ̊ apart. Fourth root: 360/4 = 90, roots occur spaced 90 ̊ apart.

37. 37 The formula k starts at 0 and goes up to n-1 This is easier than it looks. Using DeMoivre’s Theorem we get

38. 38 General Process 1.Problem must be in trig form 2.Take the n th root of r. All answers have the same value for r. 3.Divide theta by n to find the first angle. 4.Divide a full circle by n to find out how much you add to theta to get to each subsequent answer.

39. 39 Example 1. Find the 4 th root of 81 2. Divide theta by 4 to get the first angle. 3. Divide a full circle (360) by 4 to find out how far apart the answers are. 4.List the 4 answers. The only thing that changes is the angle. The number of answers equals the number of roots.

40. 40 Problems

41. 41 Homework Section 8.5 Notes Worksheet Sec. 8.5 pg. 641; # 89 – 103 odd