Basics of complex number Step-by-Step Math Problem Solver (2024)

COMPLEX NUMBERS

In the next section we will solve quadratic equations, which have a term raised to the second power (for example, x^2- 4x + 3 = 0). Solutions of quadratic equations may not be real numbers. For example, there are no real number solutions to the quadratic equation

x^2+1=0.

A set of numbers is needed that permits the solution of all quadratic equations. To get such a set of numbers, the number i is deﬁned as follows.

DEFINITION OFi i^2=-1 or i=root(-1)

Numbers of the form a + bi, where a and b are real numbers, are called complex numbers. Each real number is a complex number, since a real number a may be thought of as the complex number a + 0i. A complex number of the form a + bi, where b is nonzero, is called an imaginary number. Both the set of real numbers and the set of imaginary numbers are subsets of the set of complex numbers. (See Figure 2.3. which is an extension of Figure 1.5 in Section 1.1.) A complex number that is written in the form a + bi or a + ib is in standard form. (The forma + ib is used to simplify certain symbols such as iroot(5), since root(5)i could be too easily mistaken for root(5i))

FIGURE 2.3 Complex numbers (Real numbers are shaded.)

Example 1.

IDENTIFYING KINDS OF COMPLEX NUMBERS

The following statements identify different kinds of complex numbers

(a)-8,root(7), andPIare real numbers and complex numbers.

(b)3i,-11i,iroot(14), and5+i are imaginary numbers and complex numbers.

Example 2.

WRITING COMPLEX NUMBERS IN STANDARD FORM

The list below shows several numbers, along with the standard form of each number.

Number	Standard Form
6i	0+6i
9	-9+0i
0	0+0i
-i+2	2-i
8+iroot(3)	8+iroot(3)

Many of the solutions to quadratic equations in the next section will involve expressions such as root(-a), for a positive real number a, deﬁned as follows.

DEFINITIONOFroot(-a) Ifa>0, then

root(-a)=iroot(a)

Example 3.

WRITINGroot(-a) ASiroot(a)

Write each expression as the product of i and a real number.

(a)root(-16)=iroot(16)=4i

(b)root(-70)=iroot(70)

Products or quotients with negative radicands are simpliﬁed by ﬁrst rewriting root(-a)as iroot(a) for positive numbers a. Then the properties of real numbers can be applied, together with the fact that i^2=-1.

The rule root(c)*root(d)=root(cd) is valid only when c and d are not both negative. For example,

root((-4)(-9))=root(36)=6,

While

root(-4)*root(-9)=2i(3i)=6i^2=-6,

so that

root((-4)(-9)) is not equal toroot(-4)*root(-9).

CAUTION When working with negative radicands, be sure to use the deﬁnition root(-a)=iroot(a)before using any of the other rules for radicals.

Example 4.

FINDING PRODUCTS AND QUOTIENTS INVOLVING NEGATIVE RADICANDS

Multiply or divide as indicated.

=4-2i

The product of two complex numbers can be found by multiplying as if the numbers were binomials and using the fact that i^2=-1, as follows.

(a+bi)(c+di)

=ac+adi+bic+bidi

=ac+adi+bci+bdi^2

=ac+(ad+bc)i+bd(-1)

(a+bi)(c+di)=(ac-bd)+(ad+bc)i

Based on this result, the product of the complex numbers a + bi and c + di is deﬁned in the following way.

MULTIPLICATION OF COMPLEX NUMBERS

(a+bi)(c+di)=(ac-bd)+(ad+bc)i

This deﬁnition is not practical to use. To ﬁnd a given product, it is easier just to multiply as with binomials.

Example 7.

MULTIPLYING COMPLEX NUMBERS

Find each of the following products

(a)(2-3i)(3+4i)

=2(3)+2(4i)-3i(3)-3i(4i)

=6+8i-9i-12i^2

=6-i-12(-1) i^2=-1

=18-i

(b)(5-4i)(7-2i)

=5(7)+5(-2i)-4i(7)-4i(-2i)

=35-10i-28i+8i^2

=35-38i+8(-1)

=27-38i

(c)(6+5i)(6-5i)

=6^2-25i^2 Product of the sum and difference of two terms

=36-25(-1) i^2=-1

=36+25

=61 or 61+0i Standard form

(d)(4+3i)^2

=4^2+2(4)(3i)+(3i)^2 Square of a binomial

=16+24i+(-9)

=7+24i

Powers of i can be simpliﬁed using the facts that i^2=-1 and 1^4=1. The next example shows how this is done.

Example 8.

SIMPLIFYING POWERS OFi.

(a)i^15

Sincei^2=-1the value of a power of i is found by writing the given power as a product involving,i^2 ori^4. For example,i^3=i^2*i=(-1)*i=-i, Also,i^4=i^2*i^2=(-1)(-1)=1. Usingi^4 andi^3 to rewritei^15 gives

i^15=i^12*i^3=(i^4)^3*i^3=(1)^3(-i)=-i

(b)i^-3=i^-4*i=(i^4)^-1*1=(1)^-1*i=i

We can use the method of Example 8 to construct the following table of powers of i.

POWERS OFi

i^1=i i^5=i i^9=i

i^2=-1 i^6=-1 i^10=-1

i^3=-i i^7=-i i^11=-i

i^4=1 i^8=1 i^12=1 and so on.

Example 7(c) showed that (6 + 5i)(6 - 5i) = 61. The numbers 6 + 5i and 6 - 5idiffer only in their middle signs; for this reason these numbers are called conjugates of each other. The product of a complex number and its conjugate is always a real number.

PROPERTY OF COMPLEX CONJUGATES

For real numbers a and b:

(a+bi)(a-bi)=a^2+b^2.

Example 9.

EXAMINING CONJUGATES AND THEIR PRODUCTS

The following list shows several pairs of conjugates, together with their products.

Number	Conjugate	Product
3-i	3+i	(3-i)(3+i)=9+1=10
2+7i	2-7i	(2+7i)(2-7i)=53
-6i	6i	(-6i)(6i)=36

The conjugate of the divisor is used to ﬁnd the quotient of two complex numbers. The quotient is found by multiplying both the numerator and the denominator by the conjugate of the denominator. The result should be written in standard form.

Example 10.

DIVIDING COMPLEX NUMBERS

(a) Find(3+2i)/(5-i)

Multiply numerator and denominator by the conjugate of5-i.

(3+2i)/(5-i)

=((3+2i)(5+i))/((5-i)(5+i))

=(15+3i+10i+2i^2)/(25-i^2) Multiply.

=(13+13i)/(26) i^2=-1

=13/26+(13i)/(26) (a+bi)/c=a/c+(bi)/c

=1/2+1/2i lowest terms

To check this answer, show that

(5-i)(1/2+1/2i)=3+2i

(b)3/i

=(3(-i))/(i(-i)) -iis the conjugate ofi.

=(-3i)/(-i^2)

=(-3i)/1 -i^2=-(-1)=1

=-3i or 0-3i Standard form

Basics of complex number Step-by-Step Math Problem Solver (2024)

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