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Objectives Arithmetic Operations on Complex Numbers
Square Roots of Negative Numbers Complex Solutions of Quadratic Equations

Complex Numbers If the discriminant of a quadratic equation is negative, the equation has no real solution. For example, the equation x2 + 4 = 0 has no real solution. If we try to solve this equation, we get x2 = –4, so But 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.

Complex Numbers To make it possible to solve all quadratic equations, mathematicians invented an expanded number system, called the complex number system. First they defined the new number This means that i 2 = –1. A complex number is then a number of the form a + bi, where a and b are real numbers.

Complex Numbers Note that both the real and imaginary parts of a complex number are real numbers.

Example 1 – Complex Numbers
The following are examples of complex numbers. 3 + 4i Real part 3, imaginary part 4 Real part , imaginary part 6i Real part 0, imaginary part 6 –7 Real part –7, imaginary part 0

Arithmetic Operations on Complex Numbers

Arithmetic Operations on Complex Numbers
Complex numbers are added, subtracted, multiplied, and divided just as we would any number of the form a + b The 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)i Multiply and collect like terms I2 = –1 Combine real and imaginary parts

Arithmetic Operations on Complex Numbers
We therefore define the sum, difference, and product of complex numbers as follows.

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) i23 Solution: (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

Example 2 – Solution (b) (3 + 5i) – (4 – 2i) = (3 – 4) + [5 –(– 2)]i
cont’d (b) (3 + 5i) – (4 – 2i) = (3 – 4) + [5 –(– 2)]i = –1 + 7i (c) (3 + 5i)(4 – 2i) = [3  4 – 5 (– 2)] + [3(– 2) + 5  4]i = i (d) i23 = i22+1 = (i2)11i = (–1)11i = (–1)i = –i

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 that z  z = (a + bi)(a – bi) = a2 + b2 So the product of a complex number and its conjugate is always a nonnegative real number.

Arithmetic Operations on Complex Numbers
We use this property to divide complex numbers.

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.

Example 3 – Solution cont’d (a) The complex conjugate of 1 – 2i is 1 – 2i = 1 + 2i.

Example 3 – Solution (b) The complex conjugate of 4i is –4i.
cont’d (b) The complex conjugate of 4i is –4i. Therefore,

Square Roots of Negative Numbers

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

Example 4 – Square Roots of Negative Numbers
(c)

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.

Square Roots of Negative Numbers
For example, but so When multiplying radicals of negative numbers, express them first in the form (where r > 0) to avoid possible errors of this type.

We have already seen that if a  0, then the solutions of the quadratic equation ax2 + bx + c = 0 are If 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.

Example 6 – Quadratic Equations with Complex Solutions
Solve each equation. (a) x2 + 9 = (b) x2 + 4x + 5 = 0 Solution: (a) The equation x2 + 9 = 0 means x2 = – 9, so x = = =  3i The solutions are therefore 3i and – 3i.

Example 6 – Solution (b) By the Quadratic Formula we have x = –2  i
cont’d (b) By the Quadratic Formula we have x = –2  i So the solutions are –2 + i and –2 – i.