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# Exponents and Radicals

**CHAPTER 5****Exponents and Radicals**

**5.1 Positive Integral Exponents.**

** **In Chapter 3 we introduced the notation that

x^n=xxx...x (n factors)

where x is any real number and n is a positive integer. The number x is called the base and n the exponent or power. Computations with exponents depend on the following ﬁve basic laws.

If x and y are real numbers and m and n are positive integers, then

**E.1 **x^nx^m=x^(n+m)

**E.2 **(xy)^n=x^(n)y^(n)

**E.3 (x^n)^m=x^nm**

**E.4 **(x/y)^n=(x^n)/(y^n)

**E.5 **

The justification of these laws involves nothing more than counting the number of factors in a given expression. Law E.1 and the ﬁrst two parts of E.5 were established in Chapter 2. We will establish E.2 and E.3 below and leave E.4 and the rest of E.5 as exercises for the interested reader.

Since (xy)^n has n factors of xy, there are n factors of x and n factors of y. Using associativity and commutativity of multiplication we have

(xy)^n=(xy)(xy)...(xy)

=(xx...x)(yy...y)

=x^(n)y^(n)

In E.3 the expression (x^n)^m has m factors of x^n. Each of these factors has n factors of x, so that altogether there are nm factors of x. Thus

(x^n)^m=x^(nm)

**Example 1. **We use the laws of exponents to compute each of the following.

(a) a^2a^4=a^(2+4)=a^6 (E.1)

(b) (2x)^4=2^4x^4=16x^4 (E.2)

(c) (y^2)^5=y^(2*5)=y^10 (E.3)

(d) (z/w)^3=(z^3)/(w^3) (E.4)

(e) (a^4)/(a^2)=a^(4-2)=a^2 (E.5)

(f) (a^5)/(a^5)=1 (E.5)

(g) x^2/x^7=1/(x^(7-2))=1/x^5 (E.5)

**Example 2. **Simplify 3a^2b^3)^3(ab^2)

3a^2b^3)^3(ab^2)

=3^3(a^2)^3(b^3)^3(ab^2)

=3^3a^6b^9ab^2

=27a^(6+1)b^(9+2)

=27a^7b^11

**Example 3. **Simplify ((10xy^4)^3)/((5x^2y)^2).

((10xy^4)^3)/((5x^2y)^2)

=(10^3x^2y^(4*3))/(5^2x^(2*2)y^2)

=((2*5)^3x^3y^12)/(5^2x^4y^2)

=(2^(3)5^3x^3y^12)/(5^2x^4y^2)

=(2^(3)5^(3-2)y^(12-2))/(x^(4-3))

=(40y^10)/(x)

Let’s see how our math solver simplifies this and similar problems. Click on "Solve Similar" button to see more examples.

** **

**5.2 Integral Exponents**

** **In this section we will enlarge our set of exponents to include zero and the negative integers. We want laws E.1 through E.5 to hold for this larger set of exponents. If a!=0, then in order for a^0 to satisfy E.1, we would have

a^0a^n=a^(0+n)=a^n

Since 1 is the only real number such that 1a^n=a^n, we deﬁne

a^0=1, (a!=0)

Similarly, if n is a positive integer, then in order For a^-n to satisfy E.1, We would have

a^(-n)a^n=a^(-n+m)

=a^0

=1

Since 1/a^n is the only real number such that (1/a^n)a^n=1. we deﬁne

a^-n=1/a^n, (a!=0)

This deﬁnition gives us the following helpful rule

1/(a^-n)=1/(1/a^n)=a^n

Note that in the above deﬁnitions, when the exponent is zero or negative, a cannot be zero.

It is straightforward but tedious to show that all the laws of Section 5.1 hold for integral exponents. In fact E.5 may be written in the simpler form

**E.5**

** x^n/x^m=x^(n-m)**

In the following examples we will use the new deﬁnitions and laws of exponents to simplify the given expressions. Furthermore, we will rewrite them without zero or negative exponents.

**Example 1. **We use the above deﬁnitions to simplify each of the following.

(a) 2^0=1

(b) (-5)^0=1

(c) -5^0=-1

(d) 10^-2=1/10^2

(e) 1/(10^-3)=10^3

**Example** **2.** Simplify (u^-4v^7)(u^2v^-5).

(u^-4v^7)(u^2v^-5)

=(u^-4u^2)(v^7v^-5)

=u^(-4+2)v^(7-5)

=u^-2v^2

=(v^2)/(u^2)

**Example** **3.** Simplify (x^2y^-2z)^2(xy^4).

(x^2y^-2z)^2(xy^4)

=(x^2)^2(y^-2)^2z^2(xy^4)

=x^4y^-4z^2xy^4

=(x^4x)(y^-4y^4)z^2

=x^5y^0z^2

=x^5z^2

**Example** **4.** Simplify ((a^-2bc^3)^-2)/((a^2b^-2c)^-3).

((a^-2bc^3)^-2)/((a^2b^-2c)^-3)

=(a^4b^-2c^-6)/(a^-6b^6c^-3)

=a^(4-(-6))b^(-2-6)c^(-6-(-3))

=a^10b^-8c^-3

=(a^10)/(b^8c^3)

**Example** **5.** Simplify (a^-3-b^-3)/(a^-1-b^-1).

(a^-3-b^-3)/(a^-1-b^-1)

=(1/a^3-1/b^3)/(1/a-1/b)

=((b^3-a^3)/(a^3b^3))/((b-a)/(ab))

=((b-a)(b^2+ba+a^2))/(a^3b^3)(ab)/(b-a)

=(b^2+ba+a^2)/(a^2b^2)

Let’s see how our math solver solves this and similar problems. Click on "Solve Similar" button to see more examples.

Large and small numbers in decimal form occur in scientiﬁc work. Integral exponents can be used to write such numbers in a compact form. For example,

132,000=1.32×100,000

=1.32×10^5

and

0.0000132=1.32×0.00001

=1.32×10^-5

We see that factoring 10^5 from 132,000 moved the decimal point ﬁve places to the left, while factoring 10^-5 from 0.0000132 moved the decimal point ﬁve places to the right. In general if n is a positive integer, then factoring 10^n from a decimal number moves the decimal point n places to the left, and factoring 10^-n moves the decimal point n places to the right. We can use this idea to write any real number r in decimal form as a number s between 1 and 10, times an integral power of 10; that is,

r=s×10^n

where n is an integer. This is called scientiﬁc notation.

**Example** **6.** Express (a) 312,500, (b) 0.000115, and (c) -0.0927 in scientiﬁc notation.

(a) 312,500=3.125×10^5

(b) 0.000115=1.15×10^-4

(c) -0.0927=-9.27×10^-2

**Example** **7.** Rewrite 3.964×10^6 and (b) 7.902×10^-3in decimal form.

(a) 3.964×10^6=3,964,000

(b) 7.902×10^-3=0.007902

**Example 8. **Calculate 236000/0.0118.

236000/0.0118

=(2.36*10^5)/(1.18*10^-2

=2.00×10^7

=20,000,000

**Example 8. **Calculate (132,000,000*0.0005)/(6000).

(132,000,000*0.0005)/(6000)

=(1.32*10^8*5*10^-4)/(6*10^3)

=((1.32)*5)/(6)×(10^8x10^-4)/10^3

=1.1×10^(8-4-3)

=1.1×10

=11

**5.3 Rational Exponents**

If a and b are real numbers, n is a positive integer, and

b^n=a

then b is an nth root of a. When n = 2 We say that b is a square root, and when n = 3 we say that b is a cube root. For example, 2 and -2 are square roots of 4, since 2^2 = 4 and (-2)^2 = 4. Since (-3)^3= -27, -3 is a cube root of -27. However, -9 has no square roots since there is no real number b such that b^2 = -9. The various cases for nth roots of a real number a are as follows.

1. If n is even and a is positive, then there are two nth roots and one is the negative of the other. The positive nth root is called the principal nth root of a.

2. If n is even and a is negative, then there are no real nth roots of a.

3. If n is odd, then for all a, there is exactly one nth root of a, which is called the principal nth root.

4. If a = 0, then for all n,0 is the only nth root of 0, and thus 0 is the principal nth root of 0.

In the last section we enlarged our collection of exponents to include the integers. In this section we extend, by using nth roots, the set of exponents to include all rational numbers in such a way that the laws of exponents E.1-E.5 still hold.

For n, a positive integer, we ﬁrst deﬁne what we mean by a^(1/n). From law E.3, if m = n, then

(a^(1/n))^n=a^((1/n)-n)=a^1=a

Thus, a^(1/n) must be an nth root of a and consequently we deﬁne a^(1/n) to

be the principal nth root of a.

Next for m and n positive integers we have, from law E.3,

(a^(1/n)^m=a^((1/n)-m)=a^(m/n)

Therefore, we deﬁne

a^(m/n)=(a^(1/n))^m

When (a^(1/n))^m and (a^m)^(1/n)are both deﬁned, then they are equal since each is the principal nth root of a^m In fact (a^m)^(1/n) is the principal nth root of a^m by deﬁnition, and (a^(1/n))^m can be shown to be the principal nth root of a^m by examining each of the four cases above.

If a^(1/n) is deﬁned, then both(a^(1/n))^m and (a^m)^(1/n) exist. In particular, if a is non-negative, then all three exist and we have

a^(m/n)=(a^(1/n))^m=(a^m)^(1/n)

In the remaining sections of the chapter, unless otherwise speciﬁed, we will assume that all the variables in the expressions represent non-negative real numbers.

If m and n are integers, then we deﬁne

a^-(m/n)=1/a^(m/n)

With the above deﬁnitions for rational exponents, it can be shown that the laws E.1-E.5 hold.

**Example 1. **Simplify (a) 27^(2/3), (b) (1/16)^(5/4), and (c) 8^-(2/3)

(a) 27^(2/3)=(27^(1/3))^2=3^2=9

Note that we could also have written

(b) (27)^(2/3)=(27^2)^(1/3)=(729)^(1/3)=9

Often it is simpler to take the root ﬁrst.

(b) (1/16)^(5/4)=[(1/16)^(1/4)]^5=(1/2)^5=1/32

(c) 8^-(2/3)=1/(8^(2/3))=1/(8^(1/3)^2)=(1/2)^2=1/4

**Example 2. **Simplify (0.0025)^(3/2).

(0.0025)^(3/2)

=(25*10^-4)^(3/2)

=(25)^(3/2)(10^-4)^(3/2)

=(25^(1/2))^(3)10^(-4*3/2)

=5^3×10^-6

=125×10^-6

=1.25×10^-4

**Example 3. **Simplify x^(1/4)x^(2/3).

x^(1/4)x^(2/3)

=x^(1/4+2/3)

=x^(11/12)

**Example 4. **Simplify x^-(3/4)(x^-(1/4)+x^(3/4)).

x^-(3/4)(x^-(1/4)+x^(3/4))

=x^-(3/4)x^-(1/4)+x^-(3/4)x^(3/4)

=x^(-3/4-1/4)+x^(-3/4+3/4)

=x^-1+x^0

=1/x+1

**Example 5. **Simplify ((x^(1/2)y^(-2/3))/(x^(3/4)y^(-1/3)))^-(1/2).

((x^(1/2)y^(-2/3))/(x^(3/4)y^(-1/3)))^-(1/2)

=(x^(1/2-3/4)y^(-2/3+1/3))^-(1/2)

=(x^(-1/4)y^(-1/3))^-(1/2)

=x^(1/8)y^(1/6)

**5.4 Radicals**

** **In Section 5.3, we deﬁned x^(1/n) to be the principal nth root of x. An alternative notation is

x^(1/n)=root(n,x)

In the case n=2,root(2,x) is simply written as root(x). There are many cases when this notation will be more convenient to use. The symbol √ is called the radical, x is called the radicand, n is called the index, and the whole expression is called the radical of x of order n.

Since, from Section 5.3,

x^(n/m)=(x^n)^(1/m)=(x^(1/m))^n

for m a natural number and n an integer we have

x^(n/m)=root(m,x^n)=(root(m,x))^n

**Example 1. **Change (a) 16^(5/3) and x^(-2/3) to radical form.

(a) 16^(5/3)=root(3,16^5)=(root(3,16))^5

(b) x^(-2/3)=1/(x^(3/2))=1/(root(x^3))=1/((root(x))^3)

**Example 2. **Change (a) root(6,5^3) and (b) 1/(root(7,23^4)to exponential form.

(a) root(6,5^3)=5^(3/6)=5^(1/2)

(b) 1/(root(7,23^4))=1/(23^(4/7))=23^(-4/7)

From the deﬁnition of radical and our laws of exponents, the following basic laws for radicals are easily derived:

**R.1** root(n,a^n)=a

**R.2 **root(n,ab)=root(n,a)root(n,b)

**R.3 root(n,(a/b))=(root(n,a))/(root(n,b))**

For example, **R.2** is derived as follows:

root(n,ab)

=a^(1/n)b^(1/n)

=root(n,a)root(n,b)

The other two rules are just as easily derived.

A number of operations with radicals involve changes in form, which may be made using** R.1, R.2**, and **R3**. We use these rules to simplify the expressions in the following examples.

**Example 3. **Simplify root(4,48).

root(4,48)

=root(4,2^4*3) **(R.2)**

=root(4,2^4)root(4,3) **(R.1)**

=2root(4,3)

**Example 4. **Simplify root(28c^5d^8).

root(28c^5d^8)

=root(2^2*7c^5d^8)

=root(2^2)root((c^2)^2)root((d^4)^2)root(7c) **(R.2)**

=2c^2d^4root(7c) **(R.1)**

When the radical has a fractional radicand it is sometimes desirable to remove the denominator from under the radical sign. This is called rationalizing the denominator. This may be accomplished by multiplying both the numerator and denominator by an appropriately chosen factor.

**Example 5. **Rationalize the denominator of root(5/32).

root(5/32)

=root(5/(2^5)

=root((5*2)/(2^5*2))

=(root(10))/(root((2^3)^2)

=(root(10))/(2^3)

In the second step, the appropriate factor is 2 since this makes the denominator a perfect square.

**Example 6. **Rationalize the denominator of root(3,(9)/(2a^2b^4).

root(3,(9)/(2a^2b^4)

=root(3,(9)/(2a^2b^4)*(2^2ab^2)/(2^2ab^2)

=root(3,(9*4ab^2)/(2^3a^3b^6)

=(root(3,36ab^2))/(root(3,(2ab^2)^3)

=(root(3,36ab^2))/(2ab^2)

The factor 2^3ab^2 was chosen in order to make the denominator a perfect cube.

**Example 7. **Rationalize the denominator of (2xy^2z)/(root(4,8x^2yz^6).

(2xy^2z)/(root(4,8x^2yz^6)

=(2xzy^2root(4,(2x^2y^3z^2)))/(root(4,2^3x^2yz^6root(4,2x^2y^3z^2))

=(2xzy^2root(4,(2x^2y^3z^2)))/(root(4,2^3x^2yz^6*2x^2y^3z^2))

=(2xzy^2root(4,(2x^2y^3z^2)))/(root(4,(2xyz^2)^4)

=(2xzy^2root(4,(2x^2y^3z^2)))/((2xyz^2))

=(y*root(4,2x^2y^3z^2))/(z)

There are also expressions in which the index of the radical can be reduced, as illustrated in the following example.

**Example 8. **Simplify root(4,9u^2v^4).

root(4,9u^2v^4)

=root(4,(3uv^2)^2)

=[(3uv^2)^2]^(1/4)

=(3uv^2)^(2/4)

=(3uv^2)^(1/2)

=(3u)^(1/2)(v^2)^(1/2)

=vroot(3u)

This simpliﬁcation can be made because the exponents of 3, u, and v have the common factor 2, which is a divisor of the index 4.

An expression that involves repeated radicals may be simpliﬁed by the use of the equivalent exponential form.

**Example 9. **Simplify root(3root(3).

root(3root(3)

=(3*3^(1/2))^(1/2)

=3^(1/2)*3^(1/4)

=root(4,27)

In general, an expression involving radicals is said to be in standard form if the following three conditions are satisﬁed.

1. The radicand has no factors with exponents as great as, or greater than the index.

2. The index of the radical is as small as possible.

3. The denominator is rationalized.

Note that in Examples 3 through 9 we have simpliﬁed the given expressions by changing them to standard form.

**5.5 Addition and Subtraction of Radicals**

Certain expressions involving radicals can be added and subtracted using the distributive law. For example,

2root(5)+7root(5)-3root(5)

=(2+7-3)root(5)

=6root(5)

There are cases where the given terms in an expression are not alike, but when written in standard form common factors appear.

**Example 1.** Simplify root(50)+root(98).

root(50)+root(98)

=root(25*2)+root(49*2)

=5root(2)+7root(2)

=12root(2)

**Example 2.** Simplify root(4,32c^5)+root(72c^3)-croot(4,162c)+6croot(2c).

root(4,32c^5)+root(72c^3)-croot(4,162c)+6croot(2c)

=2croot(4,2c)+6croot(2c)-3croot(4,2c)+6croot(2)

=-croot(4,2c)+12croot(2c)

**Example 3.** Simplify root(8/x)-(root(18x))/x+7root(32x^3.

root(8/x)-(root(18x))/x+7root(32x^3

=root((2^3x)/(x^2))-(root(3^2*2x))/x+7root(2^5x^3)

=(2root(2x))/x-(3root(2x))/x+7*2^2xroot(2x)

=(-1/x+28x)root(2x)

**5.6 Multiplication and Division of Radicals**

From rules R.2 and R3 of Section 5.4, it is clear that two radicals of the same index may be multiplied or divided by carrying out the operation under the radical sign.

**Example 1.**

** **(a) root(3)root(5)=root(3*5)=root(15)

(b) (root(3,81))/(root(3,3))=root(3,81/3)=root(3,27)=3

When the radicals have different indices, change the expressions to exponential form in order to simplify.

**Example 2. **simplify root(5a)root(3,2a^2).

root(5a)root(3,2a^2)

=(5a)^(1/2)(2a^2)^(1/3)

5^(1/2)a^(1/2)2^(1/3)a^(2/3)

=5^(3/6)a^(3/6)2^(2/6)a^(4/6)

=(5^(3)a^(3)2^(2)a^4)^(1/6)

=root(6,5^(3)2^(2)a^(7)

=aroot(6,5^(3)2^(2)a)

In the above example, to write the terms under a common radical it was necessary to express the fractional exponents as equivalent fractions with denominators being their L.C.D. We use the same technique when we divide.

**Example 3. **Divide root(3,2x) by root(8x).

(3root(2x))/(root(8x))

=(root(3,2x))/(root(8x))(root(2x))/(root(2x))

=((2x)^(1/3)(2x)^(1/2))/(root(16x^2))

=((2x)^(5/6))/(4x)

=(root(6,32x^5))/(4x)

The two expressions aroot(x)+broot(y) and aroot(x)-broot(y); are called conjugates of each other. The product of these expressions is:

(a√x+b√y)(a√x-b√y)

=(a √x)^2-ab√x√y+ab√y√x-(b√y)^2

=a^2x-b^2y

For example, the conjugate of 1+2root(3) is 1-2root(3), while the conjugate of -root(2)-root(3) is -root(2)+root(3). Their respective products are

(1+2root(3))(1-2root(3))

=1-2root(3)+2root(3)-(2root(3))^2

=1-12

=-11

and

(-root(2)-root(3))(-root(2)+root(3))

=(-root(2))^2-root(2)root(3)+root(3)root(2)-(root(3))^2

=2-3

=-1

Conjugates can be used to simplify certain fractions whose denominators are of the form a√x+b√y, by removing the radicals from the denominators. As in Section 5.4 this also is called rationalizing the denominator.

**Example 4. **Simplify (root(3)+root(5))/(root(3)-root(5))by rationalizing the denominator.

(root(3)+root(5))/(root(3)-root(5))

=(root(3)+root(5))/(root(3)-root(5))*(root(3)+root(5))/(root(3)+root(5))

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

=(8+2root(15))/(-2)

=-4-root(15)

**5.7 Radicals with Negative Variables**

We recall that in the previous sections we limited our computations with nth roots to non-negative expressions. This allowed us to assume that

a^(m/n)=root(n,a^m)=(root(n,a))^m

In considering negative radicands there are certain difﬁculties. In the case that n is odd, the above equations sh']_l hold and the laws of exponents and radicals still apply. In particular, if n is odd,

root(n,-a)=-root(n,a)

This equality holds since

root(n,-a)

=root(n,(-1)a)

=root(n,-1)root(n,a)

=(-1)root(n,a)

=-root(n,a)

For example, root(3,-8)=-root(3,8)=-2. In the case that n is even, root(n,a^m) may exist while root(n,a^m) may not. For example, root((-3)^2)=root(9)=3, while (root(-3))^ 2 is not deﬁned. This is one of the reasons why it was convenient to consider only non-negative radicands.

We can, however, simplify expressions such as root((-3)^2) by using the idea of absolute value. In fact, root((-3)^2)=|-3|=3. In general, if n is even, then

root(n,x^n)=|x|

In the following examples the variables are allowed to take on negative values.

**Example 1.** Write root(4,16a^6x^4y^2) in standard form.

root(4,16a^6x^4y^2)

=root(4,2^4a^4x^4a^2y^2)

=2|a||x|root(4,a^2y^2)

=2|ax|root(4,a^2y^2)

=2|ax|root(|ay|)

Note that in reducing root(4,a^2y^2) we must use |ay| since ay is negative when aa and y have opposite sign.

**Example 2.** Simplify root(3,(-3x^4)/(16y^2)).

root(3,(-3x^4)/(16y^2))

=root(3,(-3x^4)/(2^4y^2)*(2^2y)/(2^2y))

=root(3,(-x^(3)12xy)/(2^6y^3)

=(-x(root(3,12xy)))/(4y)

**Example 3.** Simplify root(36x^2y^4z).

root(36x^2y^4z)

=root(6^2x^2y^4z)

=6|x||y^2|root(z)

=6|x|y^2root(z)

Note that, in the last line, since y^2 always non-negative we replace |y^2| by y^2.

**5.8 Complex Numbers.**

In our previous discussions of the real number system and radicals, it was pointed out that root(a) is not a real number in the case that n is even and a is negative. For example, root(-2), root(4,-8), and root(6,-3), do not represent real numbers. However, we now introduce new numbers, called complex numbers, which give these expressions meaning. Surprisingly, complex numbers have applications to problems in engineering and physics.

We introduce a new number i whose square is -1. Thus,

-1=i

If b is any positive real number, then

root(-b)

=root(b(-1))

=root(b)root(-1)

=root(b)i

Hence, the square root of any negative real number can be represented as the product of a real number and the number i. For example,

root(-16)=root(16)root(-1)=4i

and

root(-5)=root(5)root(-1)=root(5)i

The number i is called the imaginary unit and any number of the form bi, where b is a real number, is called an imaginary number.

We now consider all expressions of the form a + bi, where a and b are real numbers. These numbers are called complex numbers. The number a is called the real part and b is called the imaginary part of a + bi. For example, 2-3i is the complex number whose real part is 2 and whose imaginary part is -3. Since

a=a+0i

every real number a is a complex number. Similarly, every imaginary number bi is a complex number since

bi=0+bi

Two complex numbers a + bi and c+di are equal if and only if a=c and b = d, that is, if and only if the real parts are equal and the imaginary parts are equal.

The sum and difference of two complex numbers are deﬁned in a manner consistent with how we have performed these operations on other radicals. We have

(a+bi)+(c+di)

=(a+c)+(bi+di)

=(a+c)+(b+d)i

and

(a+bi)-(c+di)

=(a-c)+(bi-di)

=(a-c)+(b-d)i

For example,

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

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

=-1+7i

and

(2-3i)-(-2-i)

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

=(4-2i)

**Example 1. **Write each of the following in the form a+bi: (a) 5-root(-9) (b) (2+root(-8))/(2)

(a) 5-root(-9)

=5-root(9)root(-1)

=5-3i

(b) (2+root(-8))/(2)

=(2+root(8)root(-1))/(2)

=(2+2root(2i))/2

=1+root(2i)

**Example 2. **Simplify (2+root(-4))+(3+root(-25))-(-6-root(-9)).

(2+root(-4))+(3+root(-25))-(-6-root(-9))

=(-2+root(4)root(-1))+(3+root(25)root(-1))-(-6-root(9)root(-1))

=2+2i+3+5i+6+3i

=11+10i

The product of two complex numbers is deﬁned so that it conforms with the product of two binomials. We simply must remember that i^2 = -1. Thus

(a+bi)(c+di)

=a(c+di)+bi(c+di)

=ac+adi+bci+bdi^2

=ac+adi+bci-bd

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

**Example 3. **Multiply 2+3i by (3-2i).

(2+3i)(3-2i)

=6-4i+9i-6i^2

=6-4i+9i+6

=12+5i

We call the complex number a-bi the complex conjugate of the number a+bi. Therefore, a+bi is the complex conjugate a-bi. If we multiply a complex number by its conjugate we always obtain a real number since

(a+bi)(a-bi)

=a^2-abi+abi-b^2i^2

=a^2+b^2

We use this fact to develop a method for dividing complex numbers.

For example,

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

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

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

=(1+5i)/(2)

=1/2+5/2i

The process of division is performed by multiplying numerator and denominator by the complex conjugate of the denominator. In general,

(a+bi)/(c+di)

=((a+bi)(c-di))/((c+di)(c-di))

=((a+bi)(c-di))/(c^2+d^2)

=(ac+bd)/(c^2+d^2)+(bc-ad)/(c^2+d^2)i