Complex Numbers In Standard Form
COMPLEX NUMBERS
In the side by side section nosotros volition solve quadratic equations, which accept a term raised to the second power (for case, x^2- 4x + 3 = 0). Solutions of quadratic equations may not be existent numbers. For example, in that location 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 every bit follows.
DEFINITION OFii^two=-1or 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 circuitous number, since a real number a may exist thought of every bit 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 up of complex numbers. (See Figure 2.three. which is an extension of Figure ane.five in Department one.1.) A circuitous number that is written in the grade a + bi or a + ib is in standard form. (The coursea + ib is used to simplify certain symbols such as i root(v), since root(five)i could exist too hands mistaken for root(5i))
Effigy 2.3 Complex numbers (Real numbers are shaded.)
Case 1.
IDENTIFYING KINDS OF Complex NUMBERS
The following statements identify dissimilar kinds of circuitous numbers
(a)-viii,root(vii), andPIare existent numbers and complex numbers.
(b)3i,-11i,i root(xiv), andv+i are imaginary numbers and complex numbers.
Example 2.
WRITING COMPLEX NUMBERS IN STANDARD FORM
The list below shows several numbers, forth with the standard class of each number.
| Number | Standard Form |
| 6i | 0+6i |
| nine | -9+0i |
| 0 | 0+0i |
| -i+two | 2-i |
| 8+i root(three) | 8+i root(3) |
Many of the solutions to quadratic equations in the next section will involve expressions such every bit root(-a), for a positive existent number a, defined as follows.
DEFINITION OFroot(-a)Ifa>0, and then
root(-a)=i root(a)
Example 3.
WRITING root(-a) Every bit i root(a)
Write each expression every bit the product of i and a real number.
(a)root(-16)=i root(16)=4i
(b)root(-70)=i root(seventy)
Products or quotients with negative radicands are simplified past first rewriting root(-a) every bit i root(a) for positive numbers a. So the properties of real numbers can be applied, together with the fact that i^2=-ane.
The rule root(c)*root(d)=root(cd) is valid only when c and d are not both negative. For example,
root((-4)(-ix))=root(36)=6,
While
root(-4)*root(-9)=2i(3i)=6i^2=-vi,
and then that
root((-4)(-9)) is not equal toroot(-4)*root(-9).
CircumspectionWhen working with negative radicands, be sure to use the definition root(-a)=i root(a) earlier using any of the other rules for radicals.
Example 4.
FINDING PRODUCTS AND QUOTIENTS INVOLVING NEGATIVE RADICANDS
Multiply or divide equally indicated.
(a)root(-7)*root(-vii)=i root(7)*i root(7)
=i^2*(root(7))^2
=(-1)*seveni^two=-one
=-7
(b)root(-6)*root(-ten)=i root(6)*i root(x)
=i^2*root(60)
=-1*2root(15)
=-2root(xv)
(c)(root(-20))/(root(-2))=((i)root(xx))/((i)root(2))=root(twenty/two)=root(10)
(d)(root(-48))/(root(24))=((i)root(20))/(root(24))=(i)root(2)
OPERATIONS ON Complex NUMBERS Complex numbers may be added, subtracted, multiplied, and divided using the properties of real numbers, as shown by the following definitions and examples.
The sum of two complex numbers a + bi and c + di is defined every bit follows.
Improver OF Circuitous NUMBERS
(a+bi)+(c+di)=(a+c)+(b+d)i
Instance 5.
Adding Complex NUMBERS
Find each sum.
(a)(3-4i)+(-two+6i)
=[3+(-2)]+[-4+6]i
=1+2i
(b)(-nine+7i)+(3-15i)
=-vi-8i
Since (a + bi) + (0 + 0i) = a + bi for all complex numbers a + bi, the number 0 + 0i is called the additive identity for circuitous numbers. The sum of a + bi and -a-bi is 0 + 0i, so the number -a-bi is chosen the negative or additive changed of a + bi.
Using this definition of additive inverse, subtraction of circuitous numbers a + biand c + di is defined every bit
(a+bi)-(c+di)
=(a+bi)+(-c-di)
=(a-c)+(b-d)i
SUBTRACTION OF COMPLEX NUMBERS
(a+bi)-(c+di)=(a-c)+(b-d)i
Instance six.
SUBTRACTING COMPLEX NUMBERS
Subtract as indicated.
(a)(-4+3i)-(six-7i)
=(-4-six)+[three-(-vii)]i
=-10+10i
(b)(12-5i)-(eight-3i)
=(12-eight)+(-5+3)i
=4-2i
The production of 2 complex numbers can exist 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^two
=ac+(advertizing+bc)i+bd(-1)
(a+bi)(c+di)=(ac-bd)+(ad+bc)i
Based on this effect, the product of the circuitous numbers a + bi and c + di is defined in the post-obit fashion.
MULTIPLICATION OF COMPLEX NUMBERS
(a+bi)(c+di)=(ac-bd)+(advertizing+bc)i
This definition is not practical to use. To find a given product, it is easier just to multiply equally with binomials.
Example 7.
MULTIPLYING Circuitous NUMBERS
Find each of the following products
(a)(2-3i)(3+4i)
=2(3)+2(4i)-3i(three)-3i(4i)
=6+8i-9i-12i^ii
=6-i-12(-ane)i^two=-one
=18-i
(b)(5-4i)(vii-2i)
=five(7)+5(-2i)-4i(7)-4i(-2i)
=35-10i-28i+8i^2
=35-38i+8(-1)
=27-38i
(c)(vi+5i)(half-dozen-5i)
=6^2-25i^2 Production of the sum and difference of two terms
=36-25(-i)i^ii=-i
=36+25
=61 or61+0i Standard form
(d)(4+3i)^2
=iv^ii+ii(4)(3i)+(3i)^2 Foursquare of a binomial
=16+24i+(-9)
=7+24i
Powers of i can be simplified using the facts that i^two=-one and 1^4=1. The side by side example shows how this is washed.
Instance eight.
SIMPLIFYING POWERS OF i.
(a)i^15
Sincei^2=-ithe value of a power of i is found past writing the given power every bit a product involving,i^two ori^4. For instance,i^3=i^2*i=(-1)*i=-i, Also,i^4=i^2*i^2=(-one)(-1)=1. Usingi^4 andi^3 to rewritei^15 gives
i^15=i^12*i^three=(i^4)^three*i^3=(1)^3(-i)=-i
(b)i^-3=i^-iv*i=(i^4)^-i*1=(1)^-i*i=i
We can use the method of Example 8 to construct the post-obit table of powers of i.
POWERS OFi
i^1=ii^5=ii^9=i
i^ii=-1i^vi=-anei^10=-i
i^iii=-ii^7=-ii^11=-i
i^four=1i^8=1i^12=1 and then on.
Example 7(c) showed that (6 + 5i)(half dozen - 5i) = 61. The numbers 6 + 5i and 6 - 5i differ just 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 post-obit listing shows several pairs of conjugates, together with their products.
| Number | Cohabit | Product |
| 3-i | 3+i | (3-i)(3+i)=ix+1=10 |
| 2+7i | 2-7i | (ii+7i)(2-7i)=53 |
| -6i | 6i | (-6i)(6i)=36 |
The conjugate of the divisor is used to find the quotient of 2 circuitous 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.
(iii+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=-one
=13/26+(13i)/(26)(a+bi)/c=a/c+(bi)/c
=1/2+1/2i everyman terms
To bank check this answer, show that
(5-i)(i/2+1/2i)=3+2i
(b)3/i
=(3(-i))/(i(-i))-i is the conjugate ofi.
=(-3i)/(-i^ii)
=(-3i)/1-i^2=-(-1)=ane
=-3i or0-3i Standard form
Complex Numbers In Standard Form,
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