3 Objectives Arithmetic Operations on Complex Numbers Square Roots of Negative NumbersComplex Solutions of Quadratic Equations
4 Complex NumbersIf the discriminant of a quadratic equation is negative, the equation has no real solution. For example, the equationx2 + 4 = 0has no real solution. If we try to solve this equation, we get x2 = –4, soBut this is impossible, since the square of any real number is positive. [For example,(–2)2 = 4, a positive number.]Thus, negative numbers don’t have real square roots.
5 Complex NumbersTo make it possible to solve all quadratic equations, mathematicians invented an expanded number system, called the complex number system.First they defined the new numberThis means that i 2 = –1.A complex number is then a number of the form a + bi, where a and b are real numbers.
6 Complex NumbersNote that both the real and imaginary parts of a complex number are real numbers.
7 Example 1 – Complex Numbers The following are examples of complex numbers.3 + 4i Real part 3, imaginary part 4Real part , imaginary part6i Real part 0, imaginary part 6–7 Real part –7, imaginary part 0
9 Arithmetic Operations on Complex Numbers Complex numbers are added, subtracted, multiplied, and divided just as we would any number of the form a + bThe only difference that we need to keep in mind is that i2 = –1. Thus, the following calculations are valid.(a + bi)(c + di) = ac + (ad + bc)i + bdi2= ac + (ad + bc)i + bd(–1)= (ac – bd) + (ad + bc)iMultiply and collect like termsI2 = –1Combine real and imaginary parts
10 Arithmetic Operations on Complex Numbers We therefore define the sum, difference, and product of complex numbers as follows.
11 Example 2 – Adding, Subtracting, and Multiplying Complex Numbers Express the following in the form a + bi.(a) (3 + 5i) + (4 – 2i) (b) (3 + 5i) – (4 – 2i)(c) (3 + 5i)(4 – 2i) (d) i23Solution: (a) According to the definition, we add the real parts and we add the imaginary parts.(3 + 5i) + (4 – 2i) = (3 + 4) + (5 – 2)i= 7 + 3i
13 Arithmetic Operations on Complex Numbers Division of complex numbers is much like rationalizing the denominator of a radical expression.For the complex number z = a + bi we define its complex conjugate to be z = a – bi.Note thatz z = (a + bi)(a – bi) = a2 + b2So the product of a complex number and its conjugate is always a nonnegative real number.
14 Arithmetic Operations on Complex Numbers We use this property to divide complex numbers.
15 Example 3 – Dividing Complex Numbers Express the following in the form a + bi. (a)(b)Solution: We multiply both the numerator and denominator by the complex conjugate of the denominator to make the new denominator a real number.
16 Example 3 – Solutioncont’d(a) The complex conjugate of 1 – 2i is 1 – 2i = 1 + 2i.
17 Example 3 – Solution (b) The complex conjugate of 4i is –4i. cont’d(b) The complex conjugate of 4i is –4i.Therefore,
19 Square Roots of Negative Numbers Just as every positive real number r has two square roots ( and – ), every negative number has two square roots as well.If –r is a negative number, then its square roots are i , because = i2r and = (–1)2i2r = –r.We usually write instead of to avoid confusion with
20 Example 4 – Square Roots of Negative Numbers (c)
21 Square Roots of Negative Numbers Special care must be taken in performing calculations that involve square roots of negative numbers.Although when a and b are positive, this is not true when both are negative.
22 Square Roots of Negative Numbers For example,butsoWhen multiplying radicals of negative numbers, express them first in the form (where r > 0) to avoid possible errors of this type.
24 Complex Solutions of Quadratic Equations We have already seen that if a 0, then the solutions of the quadratic equation ax2 + bx + c = 0 areIf b2 – 4ac < 0, then the equation has no real solution.But in the complex number system, this equation will always have solutions, because negative numbers have square roots in this expanded setting.
25 Example 6 – Quadratic Equations with Complex Solutions Solve each equation.(a) x2 + 9 = (b) x2 + 4x + 5 = 0Solution: (a) The equation x2 + 9 = 0 means x2 = – 9, sox === 3iThe solutions are therefore 3i and – 3i.
26 Example 6 – Solution (b) By the Quadratic Formula we have x = –2 i cont’d(b) By the Quadratic Formula we havex= –2 iSo the solutions are –2 + i and –2 – i.
27 Complex Solutions of Quadratic Equations We see from Example 6 that if a quadratic equation with real coefficients has complex solutions, then these solutions are complex conjugates of each other.So if a + bi is a solution, then a – bi is also a solution.