Module on Complex Numbers

1. Module on Complex Numbers Lesson 1: Basic Understanding and Operations of Complex Numbers (slides ?-?) Lesson 2: Geometric understanding of addition and subtraction (slides ?-?) Lesson 3: Applications (slides ?-?)

2. Click on the picture to watch the video Video Annenberg

3. Lesson 1: Basic Understanding and Operations of Complex Numbers Introduction: Why do we need new numbers? • The hardest thing about working with complex numbers is understanding why you might want to. Before introducing complex numbers, let's backup and look at simpler examples of the need to deal with new numbers. • If you are like most people, initially number meant whole number, 0,1,2,3,... Whole numbers make sense. They provide a way to answer questions of the form "How many ... ?" You also learned about the operations of addition and subtraction, and you found that while subtraction is a perfectly good operation, some subtraction problems, like 3 - 5, don't have answers if we only work with whole numbers. 3

4. Then you find that if you are willing to work with integers, ...,-2, -1, 0, 1, 2, ..., then all subtraction problems do have answers! Furthermore, by considering examples such as temperature scales, you see that negative numbers often make sense. • Now that we have fixed subtraction we will deal with division. Some, in fact most, division problems do not have answers that are integers. For example, 3 ÷ 2 is not an integer. We need new numbers! Now we have rational numbers (fractions). • There is more to this story. There are problems with square roots and other operations, but we will not get into that here. The point is that you have had to expand your idea of number on several occasions, and now we are going to do that again. 4

5. The "problem" that leads to complex numbers concerns solutions of equations. • What if we want to find a number that when multiplied it by itself equals -1?" • Or can you solve the equation x2 + 4 = 0 for x? **************************************************** • The need to extend the real numbers was prompted by the desire to solve problems like the following, which appears in the 1545 book Ars Magna by Cardana: • “Divide 10 into two parts whose product is 40.” • To solve this problem, Cardano needed to solve the equation x(10-x)=40, which is equivalent to x2-10x+40=0. • The solution of the equation requires to be a number. But -15 does not have a square root that is a real number. To overcome this barrier, a new number system was invented. 5

6. In the real number system, the square root of a negative number does not exist. That's because there are no real numbers whose squares are negative. • However, you can take the square root of a negative number if you are willing to use a non-real number to do it. This new number was invented (discovered?) more than 400 years ago. It was called "i", standing for "imaginary", because i wasn't "real". • Imaginary number: 6

7. But i already squares to –1. So it cannot also square to 1. This points out an important detail: when dealing with imaginary numbers, you gain something (the ability to deal with negatives inside square roots), but you also lose something (some of the convenient rules you have when dealing with square roots). Lacking these rules, we make the following definition: • If x is a positive real number, then • Examples: • You can check our answers on a calculator which permits complex number arithmetic. The TI-84 is one such calculator, and we will use it throughout this module. After turning it on, press the MODE key. Move down to REAL and over to “a + bi”. With the cursor blinking over “a + bi”, press ENTER to save this complex mode of arithmetic. Now go back to the home screen by pressing 2ND QUIT. You are ready to take the square roots of negative numbers. Try it on the examples above! 7

8. Powers of i: • We have seen that . It then follows that 8

9. Notice that the powers of i cycle around four numbers: i, -1, -i, 1 • If the exponent is a multiple of 4, the power equals 1. For example, • To calculate any high power of i, you can convert it to a lower power by taking the closest multiple of 4 which is no larger than the exponent and subtracting this multiple from the exponent. For example, • You can confirm these answers by calculator: The “i” key is found in the middle of the bottom row. To compute i to the power 27, press 2ND i ^27 ENTER. Your answer should be –i, but perhaps you instead got something like -3E-13 - i. What is going on here? The real part of this result is -3 times 10 to the power -13, which is within machine roundoff error of zero. So, you may assume that the calculator has given an approximation to 0 – i; in other words, -i. 9

10. Pause and practice-1 • Simplify • First compute these by the rules above, then check answers on your calculator. Answer: (a) –i (b) -1 (c) i 10

11. Complex numbers: • What if we want to combine a real number with an imaginary number? We could say that 3 + 4i is a number but it has more parts to it than a normal number. We call it a complex number. • Every complex number can be written in the form a + bi, where a and b are real numbers, called the real part and the imaginary part of the complex number, respectively. • For example, 2 + 3i is a complex number, with real part 2 and imaginary part 3. 11

12. It's as though our imaginary number 'i' isn't on the number line, but we must be able to put it somewhere. And what about 2i and 3i and -7i? We must be able to put them all somewhere. Why don't we make our imaginary number line perpendicular to the real number line through the origin. then not only will we have a place for imaginary numbers like 5i and -3i but also for complex numbers like 2+4i and -2 - 5i. Its like we now have a visual way for looking at complex numbers. • A complex number can be visually represented as an ordered pair of numbers specifying a point in the xy-plane. This point is the tip of a vector emanating from the origin, and this vector also represents the complex number.

13. Below is a graph of the vector representing the number 2 + 3i . Note that the tip of the vector has coordinates (2, 3). In general, any complex number a + bi can be plotted in the xy-plane (also called the complex plane) as the point having coordinates (a, b). This point is the tip of a vector emanating from the origin. Important special cases: a pure real number is plotted on the x-axis, and a pure imaginary number is plotted on the y-axis. 13

14. Arithmetic of complex numbers: • Given two complex numbers, we now define how to add, subtract, multiply, and divide them. We want to do this in a natural way so that the usual rules for arithmetic of real numbers continue to be valid for complex numbers. In particular, we want addition and multiplication to be commutative, associative, and distributive. Since the complex number a + bi looks a bit like the linear polynomial a + bx, let’s use our knowledge of polynomials as motivation. So, to add, subtract, or multiply complex numbers, we simply combine like terms – as is seen in the following examples: 14

15. Operations: • Addition example: • Subtraction example: APPLET-1 15

16. Operations: • Multiplication example: • Note that we replaced by -1 in the above calculation. APPLET-2 16

17. Pause and practice-2 • Add, subtract, and multiply the complex numbers and . • First do these problems by hand, then check your results by calculator. 17

18. What about division? • If the denominator is pure imaginary, we can multiple by to eliminate the iin the denominator, as is seen in this example: • This was simple enough, but what if you have something more complicated, such as • In order to eliminate the i in the denominator, we make use of the conjugate. 18

19. The conjugate of a complex number a + bi: • The conjugate of a complex number a + bi is the same number, but with the opposite sign in the middle: a – bi. For example, the conjugate of 3 + 4i is 3 – 4i. • The multiplication by conjugates produces a sum of squares. You should pause to verify this fact: • This is similar to the more familiar difference of squares: 19

20. Division example: • Note that in the first step we multiplied both numerator and denominator by the conjugate of the denominator. • In the last step, note how the fraction was split into two pieces. This is because, technically speaking, a complex number is expressed as a sum of two parts: a + bi. • To check this example on your calculator, enter (5+2i)÷(3+4i) to get .92 - .56i. If you prefer fractions over decimals, press MATH FRAC ENTER. 20

21. Pause and Practice -3 • Divide: (3 + 4i) ÷ (5 – 2i) by hand, and then check your answer by calculator. 21

22. Modulus (or, Absolute Value): • We have seen how complex numbers can be represented by points in the complex plane. Unlike the real numbers, there is not a natural way to order them. For example, 3 < 4, but how would you compare 3 + 2i and 4 + i? We do this by computing the distance each point is from the origin. Using the distance formula, • The modulus (or, absolute value ) of a complex number a + bi is defined to be its distance to the origin, and is denoted by • Note that this definition agrees with the definition of absolute value of a real number. 22

23. In our example, we would say that the modulus of 3 + 2i is less than the modulus of 4 + i. • You can check your answers on the calculator using abs(a+ib), as seen in the following example: • Select MATH NUM abs(3+2i) ENTER, giving a decimal approximation to the square root of 13. Since this number is irrational, you cannot convert it to a fraction using FRAC. 23

24. Pause and practice-4 • For each of the following complex numbers, plot the complex number as a point in the complex plane, connect the point to the origin by a segment, and compute the modulus (first by hand, then by calculator) to obtain the length of this segment. (a) 3 + 4i (b) -5 + 12i (c) 8 – 6i 24

25. Lesson 1 - Quiz Try to answer each question twice, without a calculator and then with a calculator. 1. 2. 25

26. Lesson 1 - Quiz 3. Draw the vector representing 2-i . 4. 5. 26

27. Lesson 1 - Quiz 6. 7. Answers: 1(b), 2(a), 4(c), 5(a), 6(b), 7(d) 27

28. Lesson 2: Geometric understanding of addition and subtraction Introduction: • In this discussion, it will be convenient to denote a complex number by a single variable name. • ?? 28

29. Let w = a + bi and z = c + di. If O denotes the origin, then the following four points form vertices of a parallelogram: O, w, z, and w + z. • For example, if w = 2 + i and z = 1 + 2i, then w + z = 3 + 3i. It is easy to see that (0, 0), (2, 1), (1, 2), and (3, 3) form vertices of a parallelogram. The points (0, 0) and (3, 3) are endpoints of one diagonal, while (2, 1) and (1, 2) are endpoints of the other diagonal. • In general, O and w + z are endpoints of one diagonal, while w and z are endpoints of the other diagonal. • Please click the button to ??. w+z z w Applet for addition 29

30. Pause and practice-5 • Sketch the parallelogram formed by the complex numbers w = 1 + 2i , z = -2 + I , origin and w+z. Is this a rectangle? a square? • Also confirm it with the applet Applet for addition 30

31. What about subtraction? • Once again, a parallelogram is formed: by the four points: O, w, z, and w – z. However, in this case O and w are endpoints of one diagonal, while z and w - z are endpoints of the other diagonal. • Try with the applet Applet for subtraction 31

32. Pause and practice-5 • O and w are endpoints of one diagonal, while z and w - z are endpoints of the other diagonal. • Confirm this fact in the case where w = 1 + 2i and z = -2 + i. • And check with the applet Applet for subtraction

33. Complex numbers in polar form • In order to gain a geometric understanding of multiplication, it will be convenient to represent points in the complex plane using polar coordinates. Suppose that z = a + bi. Rather than represent this point using rectangular xy-coordinates, we can use polar coordinates (r, Ө). The variable r represents the distance from point z to the origin, and Ө is the angle (measured counterclockwise) the vector z makes with the positive x-axis. 33

34. Using the definitions of the trigonometric functions sine and cosine, we have • Therefore, z can be written • The expression arises so frequently that it is customary to abbreviate it as “cis ”. Then, z = rcis is called the polar form of a complex number. • Since the distance to the origin is represented by the modulus of z, we have . The angle Ө is called the argument of z. Note that the argument can have many possible values, but any two of them must differ by a multiple of 360 degrees. 34

35. There is an interesting connection between the polar form of a complex number and the real function , which is the inverse function of the natural logarithm. In the 18th century, Leonhard Euler discovered that , provided that is measured in radians. Thus, and so the polar form of a complex number can be written as • The polar form is available on the TI-84 calculator. Simply press MODE, then move the cursor down to REAL and over to “re^ ”, then press ENTER. The value of is expressed in radians or degrees, depending on the MODE setting. 35

36. Example: Find the polar form of z = 1 + i. • The vector represented by z is the segment from the origin to the point (1, 1). By drawing a sketch, one easily sees that the argument is 45 degrees ( multiples of 360º). In cases which are not clear, one can use the formula . In our example, both x and y have value 1. • To obtain the modulus, . Thus, the polar form is • You can check this answer on your calculator by typing: 1 + i ENTER. Of course, your MODE setting should already be set to polar: “re^ ” 36

37. Example: Find the standard form of z = 2 cis 150º. • We make use of the formulas • This gives • Thus, • We now check this answer on the calculator. Since there is no “cis” key, we make use of the key 2ND . Also, recall that Euler’s formula is valid only in radians. So, we first must convert 150 degrees to radians and enter our expression as ENTER. This gives the desired result, but with an approximation to . Note that 2e^(150*i) ENTER gives a very different (and wrong!) answer. 37

38. Pause and practice-6 • Try these by hand, and then check with your calculator. (a) Find the polar form of z = 4 – 3i. (b) Find the standard form of z = 2 cis 225º. 38

39. Geometric understanding of multiplication • We now show that the product of two complex numbers has modulusequal to the product of the individual moduli and argument equal to the sum of the individual arguments. • Let w = |w| cisA and z = |z| cisB. Then, wz = (|w| cisA)( |z| cisB) = |w||z|(cosA + i sin A)(cosB + i sin B) = |w||z|(cosAcosB – sin A sin B + i sin AcosB + icosA sin B) 39

40. Using the trigonometric angle sum formulas, this last expression can be written as = |w||z|(cos(A + B) + i sin(A + B)). Thus, the polar form of wz is wz= |w||z|(cos(A + B) + i sin(A + B)), which allows us to identify |w||z|as the modulus of wz, and A + B as the argument of wz. 40

41. Pause and practice-7 • Let w = 2i, and let z = 1 + i. (a) Find the product wz. (b) Find the modulus of each of w, z, and wz, and then try to decide how these three moduli are related to each other. Relate your answers to the polar forms of all three numbers (Check their polar forms on your calculator.) (c) Plot the three points w, z, and wz in the complex plane, and connect each point to the origin with a line segment. Try to decide how the angles between the positive real axis and these three lines are related to one another. Relate your answers to the polar forms of all three numbers (Check their polar forms on your calculator.) 41

42. De Moivre’s Theorem: • We can use the polar form to square a complex number: where A is the argument of z. where A is the argument of z. As you can see, we have squared the modulus and doubled the argument. Continuing to multiply a complex number by itself, we get a result known as de Moivre’s Theorem: • Example: Simplify . The modulus of 1 + i is , and we can choose A = 45 degrees. Thus, = = = 16 cis(360) = 16 42

43. Pause and practice-8 • Simplify • After computing by hand as in the previous example, check your answer by calculator. 43

44. Roots of complex numbers: • We now use de Moivre’s Theorem to find roots: suppose that and we would like to solve for w in terms of z. Writing the polar forms of both sides of this equation, we have or . • Equating the moduli of each side, we have , hence . • Since the arguments of each side are equal (or, differ by a multiple of 360 degrees), there is an integer k so that , hence . 44

45. We conclude that the polar form of the nth root of z is • Since k can be any integer, it may appear that there are infinitely many roots. But since cosine and sine have period equal to 360º, we get different roots only for k = 0, 1, 2, …, n-1: when k = n, we get the same “cis” value that we got for k = 0; when k = n+1, we get the same “cis” value that we got for k = 1, etc. We conclude that a nonzero complex number has exactly nnth roots. 45

46. Example: Find the three cube roots of z = -8. • For the modulus, we have . For the argument, we have A = 180º, so that for k = 0, 1, 2, the three values of are 60, 180, and 300. Thus, the three cube roots of -8 are: 2 cis 60, 2 cis 180, and 2 cis 300, which are simplified as: • The first of these three is the principal root, the one corresponding to k = 0. • You can try to check these answers on your calculator, but don’t expect to get all three answers. The TI-84 seems to give a real answer, is there is one. If none of the roots are real, the calculator gives the principal root. Let us check this for the cube root of -8. Entering (-8)^(1/3) produces an answer of -2, which is the real root but not the principal root. 46

47. Pause and practice-9 • Find the four 4th roots of z = -16, and check you answer on a calculator. You should find that there are no real roots, so the calculator displays the principal root: 47

48. Lesson 2 - Quiz 1. Let and . Sketch the parallelogram determined by Also sketch the parallelogram determined by . 2. Find the polar from of . (a) 2 cis 120° (b) 2 cis 150º (c) 2 cis 210 3. Find the standard form of z = 4 cis 135°. (a) (b) (c)

49. 4. Sketch the vectors representing the two complex numbers 5 cis 45º and 3 cis 90°. Sketch and write the product in polar form. (a) 15 cis 135º (b) 8 cis 135° (c) 15 cis 4050º 5. Explain why 6. The following are the cube roots of i. Which one is principal? (a) (b) (c) Answers: 2(c), 3(b), 4(a), 6(b)

50. Lesson 3: Applications Solving quadratic equations • Remember that the quadratic formula solves the quadratic equation "ax2 + bx + c = 0" for the values of x (called zeros of the equation). They called zeros because they are the values of x that make y = 0, in the quadratic function y = ax2 + bx + c. • Recall that a quadratic equation has two, one, or no real zeros, depending of the sign of the discriminant , which appears under the square root in the quadratic formula: 50