1. 1 Copyright © Cengage Learning. All rights reserved. 2. Equations and Inequalities 2.4 Complex Numbers

2. 2 Complex Numbers Complex numbers are needed to find solutions of equations that cannot be solved using only the set of real numbers. The following chart illustrates several simple quadratic equations and the types of numbers required for solutions.

3. 3 Complex Numbers The solutions of the first three equations in the chart are in; however, since squares of real numbers are never negative, does not contain the solutions of x 2 = –9. To solve this equation, we need the complex number system which contains both and numbers whose squares are negative. We begin by introducing the imaginary unit, denoted by i, which has the following properties.

4. 4 Complex Numbers Because its square is negative, the letter i does not represent a real number. It is a new mathematical entity that will enable us to obtain. Since i, together with, is to be contained in, we must consider products of the form bi for a real number b and also expressions of the form a + bi for real numbers a and b.

5. 5 Complex Numbers The next chart provides definitions we shall use.

6. 6 Example 1 – Addition and multiplication of complex numbers Express in the form a + bi, where a and b are real numbers: (a) (3 + 4i) + (2 + 5i) (b) (3 + 4i)(2 + 5i) Solution: (a) (3 + 4i) + (2 + 5i) = (3 + 2) + (4 + 5)i = 5 + 9i (b) (3 + 4i)(2 + 5i) = (3 + 4i)(2) + (3 + 4i)(5i) = 6 + 8i + 15i + 20i 2

7. 7 Example 1 – Solution = 6 + 23i + 20(–1) = –14 + 23i cont’d

8. 8 Complex Numbers The set of real numbers may be identified with the set of complex numbers of the form a + 0i. It is also convenient to denote the complex number 0 + bi by bi. Thus, (a + 0i) + (0 + bi) = (a + 0) + (0 + b)i = a + bi. Hence, we may regard a + bi as the sum of two complex numbers a and bi (that is, a + 0i and 0 + bi). For the complex number a + bi, we call a the real part and b the imaginary part.

9. 9 Example 2 – Equality of complex numbers Find the values of x and y, where x and y are real numbers: (2x – 4) + 9i = 8 + 3yi Solution: We begin by equating the real parts and the imaginary parts of each side of the equation: 2x – 4 = 8 and 9 = 3y Since 2x – 4 = 8, 2x = 12 and x = 6. Since 9 = 3y, y = 3. The values of x and y that make the complex numbers equal are x = 6 and y = 3

10. 10 Complex Numbers With complex numbers, we are now able to solve an equation such as x 2 = –9. Specifically, since (3i) (3i) = 3 2 i 2 = 9(–1) = –9, we see that one solution is 3i and another is –3i.

11. 11 Complex Numbers In the next chart we define the difference of complex numbers and multiplication of a complex number by a real number. If we are asked to write an expression in the form a + bi, the form a – di is acceptable, since a – di = a + (–d)i.

12. 12 Example 3 – Operations with complex numbers Express in the form a + bi, where a and b are real numbers: (a) 4(2 + 5i ) – (3 – 4i ) (b) (4 – 3i )(2 + i ) (c) i(3 – 2i ) 2 (d) i 51 (e) i –13 Solution: (a) 4(2 + 5i ) – (3 – 4i ) = 8 + 20i – 3 + 4i = 5 + 24i (b) (4 – 3i )(2 + i ) = 8 – 6i + 4i – 3i 2 = 11 – 2i

13. 13 Example 3 – Solution (c) i(3 – 2i ) 2 = i(9 – 12i + 4i 2 ) = i(5 – 12i ) = 5i – 12i 2 = 12 + 5i (d) Taking successive powers of i, we obtain i 1 = i, i 2 = –1, i 3 = –i, i 4 = 1, and then the cycle starts over: i 5 = i, i 6 = i 2 = –1, and so on. cont’d

14. 14 Example 3 – Solution In particular, i 51 = i 48 i 3 = (i 4 ) 12 i 3 = (1) 12 i 3 = (1)(–i) = –i. cont’d

15. 15 Example 3 – Solution (e) In general, multiply i –a by i b, where a  b  a + 3 and b is a multiple of 4 (so that i b = 1). For i –13, choose b = 16. i –13  i 16 = i 3 = –i cont’d

16. 16 Complex Numbers The following concept has important uses in working with complex numbers. Since a – bi = a + (–bi), it follows that the conjugate of a – bi is a – (–bi ) = a + bi. Therefore, a + bi and a – bi are conjugates of each other.

17. 17 Complex Numbers Illustration: Conjugates Complex number Conjugate 5 + 7i 5 – 7i 5 – 7i 5 + 7i 4i –4i 3 3

18. 18 Complex Numbers The following two properties are consequences of the definitions of the sum and the product of complex numbers. Note that the sum and the product of a complex number and its conjugate are real numbers. Conjugates are useful for finding the multiplicative inverse of a + bi, 1/(a + bi ), or for simplifying the quotient of two complex numbers.

19. 19 Complex Numbers As illustrated in the next example, we may think of these types of simplifications as merely rationalizing the denominator, since we are multiplying the quotient by the conjugate of the denominator divided by itself.

20. 20 Example 4 – Quotients of complex numbers Express in the form a + bi, where a and b are real numbers Solution:

21. 21 Example 4 – Solution cont’d

22. 22 Complex Numbers If p is a positive real number, then the equation x 2 = –p has solutions in. One solution is, since Similarly, is also a solution.

23. 23 Complex Numbers The definition of in the next chart is motivated by = –r for r > 0. When using this definition, take care not to write when is intended. The radical sign must be used with caution when the radicand is negative.

24. 24 Complex Numbers For example, the formula, which holds for positive real numbers, is not true when a and b are both negative, as shown below: But

25. 25 Complex Numbers If only one of a or b is negative, then In general, we shall not apply laws of radicals if radicands are negative. Instead, we shall change the form of radicals before performing any operations, as illustrated in the next example.

26. 26 Example 5 – Working with square roots of negative numbers Express in the form a + bi, where a and b are real numbers: Solution: First we use the definition, and then we simplify: = (5 – 3i )(–1 + 2i ) = –5 + 10i + 3i – 6i 2 = –5 + 13i + 6 = 1 + 13i

27. 27 Example 7 – An equation with complex solutions Solve the equation x 3 – 1 = 0. Solution: Using the difference of two cubes factoring formula with a = x and b = 1, we write x 3 – 1 = 0 as (x – 1)(x 2 + x + 1) = 0.

28. 28 Example 7 – Solution Setting each factor equal to zero and solving the resulting equations, we obtain the solutions or, equivalently, Since the number 1 is called the unit real number and the given equation may be written as x 3 = 1, we call these three solutions the cube roots of unity. cont’d