1 Properties of real numbers (Section P.1 , page 12)
2 Special products (Section P.3 , pages 34 and 37)
3 Factoring (Section P.4 )
In the early days of the study of alternate current (AC) circuits, scientists concluded that AC circuits were somehow different from the battery-powered direct current (DC) circuits. However, both types of circuits obey the same physical and mathematical laws. In 1893, Charles Steinmetz provided the breakthrough in understanding AC circuits. He explained that in AC circuits, the voltage, current, and resistance (called impedance in AC circuits) were not scalars, but alternate in direction, and possessed frequency and phase shift that must be taken into account. He advocated the use of the polar form of complex numbers to provide a convenient method of symbolically denoting the magnitude, frequency, and phase shift simultaneously for the AC circuit quantities voltage, current, and impedance. Steinmetz eventually became known as “the wizard who generated electricity from the square root of minus one.” In Example 7, we use complex numbers to compute the total impedance in an AC circuit.
1 Define complex numbers.
Because the square of a real number is nonnegative (that is, x2≥0
With the introduction of the number i, the equation x2=−1
A complex number z written in the form a+bi
We can express the square root of any negative number as a product of a real number and i.
We usually write i√b
Here is a visual reminder of the parts of a complex number.
Identify the real and imaginary parts of each complex number.
2+5i
7−12i
3i.
−9
0
3+√−25
To identify the real and imaginary parts, we express each number in the standard form a+bi.
2+5i
7−12i=7+(−12)i;
3i=0+3i;
−9=−9+0i;
0=0+0i;
3+√−25=3+(√25)i=3+5i;
Identify the real and imaginary parts of each complex number.
−1+2i
−13−6i
8
2 Add and subtract complex numbers.
The standard form of a complex number a+bi
To add or subtract complex numbers, add or subtract the real parts and add or subtract the imaginary parts. Write the sum or difference in standard form, a+bi.
Write the sum or difference of two complex numbers in standard form.
(3+7i)+(2−4i)
(5+9i)−(6−8i)
(2+√−9)−(−2+√−4)
(3+7i)+(2−4i)=(3+2)+[7+(−4)]i=5+3i
(5+9i)−(6−8i)=(5−6)+[9−(−8)]i=−1+17i
(2+√−9)−(−2+√−4)=(2+3i)−(−2+2i)√9=3i, √−4=2i=(2−(−2))+(3−2)i=4+i
Write the following complex numbers in standard form.
(1−4i)+(3+2i)
(4+3i)−(5−i)
(3−√−9)−(5−√−64)
3 Multiply complex numbers.
We multiply complex numbers by using FOIL (as we do with binomials; see page 35) and then replace i2
Write the following products in standard form.
(3−5i)(2+7i)
−2i(5−9i)
FOIL(3−5i)(2+7i)=6+21i−10i−35i2=6+11i+35Because i2=−1, −35i2=35.=41+11iCombine terms.
−2i(5−9i)=−10i+18i2Distributive property=−10i−18Because i2=−1, 18i2=−18.=−18−10i
Write the following products in standard form.
(2−6i)(1+4i)
−3i(7−5i)
Recall from algebra that if a and b are positive real numbers, then
However, this property is not true for all complex numbers. For example,
but
Thus,
When performing multiplication (or division) involving square roots of negative numbers (say, √−b
Perform the indicated operation and write the result in standard form.
√−2√−8
√−3(2+√−3)
(−2+√−3)2
(3+√−2)(1+√−32)
√−2√−8=i√2⋅i√8=i2√2⋅8=i2√16=(−1)(4)=−4
√−3(2+√−3)=i√3(2+i√3)√−3=i√3=2i√3+i2√9Distributive property=2i√3+(−1)(3)i2=−1,√9=3=−3+2i√3Standard form
(−2+√−3)2=(−2+i√3)2√−3=i√3=(−2)2+2(−2)(i√3)+(i√3)2(a+b)2=a2+2ab+b2=4−4i√3+3i2(i√3)2=(i√3)(i√3)=3i2=4−4i√3+3(−1)i2=−1=1−4i√3Simplify.
(3+√−2)(1+√−32)=(3+i√2)(1+i√32)√−2=i√2, √−32=i√32=3+3i√32+i√2+(i√2)(i√32)FOIL=3+3i(4√2)+i√2+i2√64√32=√16⋅2=4√2=3+12i√2+i√2+(−1)(8)i2=−1=−5+13√2iAdd.
Perform the indicated operation and write the result in standard form.
(−3+√−4)2
(5+√−2)(4+√−8)
4 Divide complex numbers.
To perform the division of complex numbers, it is helpful to learn about the conjugate of a complex number.
Note that the conjugate ˉz
Find the product zˉz
z=2+5i
z=1−3i
If z=2+5i,
If z=1−3i,
Find the product zˉz
1+6i
−2i
The results in Example 5 correctly suggest the following theorem.
Notice that the product of a complex number and its conjugate is always a real number.
To write the reciprocal of a nonzero complex number or the quotient of two complex numbers in the form a+bi,
Write the following quotients in standard form.
12+i
4+√−252−√−9
The denominator is 2+i,
We write √−25=5i
Write the following quotients in standard form.
21−i
−3i4+√−25
In a parallel circuit, the total impedance Zt
Find Zt
We first calculate Z1+Z2
Repeat Example 7 assuming that Z1=1+2i and Z2=2−3i.
5 Solve quadratic equations having complex solutions.
The methods of solving quadratic equations introduced in the previous section are also applicable to solving quadratic equations with complex solutions.
Solve each equation.
x2+4=0
x2+2x+2=0
Use the square root method.
The solution set is {−2i,2i}. You should check these solutions.
x2+2x+2=1⋅x2+2x+2=0
The solution set is {−1−i, −1+i}. You should check these solutions.
Solve.
4x2+9=0
x2=4x−13
Recall that in the quadratic formula
the quantity b2−4ac under the radical sign is called the discriminant. If a, b, and c are real numbers (not necessarily integers) and a≠0, the discriminant reveals the type of solutions of the equation ax2+bx+c=0, as shown in the table below.
Discriminant | Description of Solutions |
---|---|
b2−4ac>0 | There are two unequal real solutions. |
b2−4ac=0 | There is one real solution. |
b2−4ac<0 | There are two nonreal complex solutions, and they are conjugates. |
Use the discriminant to determine the number and type of solutions of each quadratic equation.
x2−4x+2=0
2t2+2t+19=0
4x2+4x+1=0
Equation | b2−4ac | Conclusion |
---|---|---|
|
(−4)2−4(1)(2)=8>0 | Two unequal real solutions |
|
(2)2−4(2)(19)=−148<0 | Two nonreal complex solutions |
|
(4)2−4(4)(1)=0 | Exactly one real solution |
Determine the number and type of solutions.
9x2−6x+1=0
x2−5x+3=0
2x2−3x+4=0.
We define i=_ so that i2=_.
A complex number in the form a+bi is said to be in .
For b>0, √−b=_.
The conjugate of a+bi is , and the conjugate of a−bi is .
True or False. The product of a complex number and its conjugate is a real number.
True or False. Division by a nonzero complex number z is accomplished by multiplying the numerator and denominator by ˉz.
True or False. A pure imaginary number does not have a conjugate.
True or False. The standard form of 1i is a+bi where a=1 and b=−1.
In Exercises 9–32, perform each operation and write the result in the standard form a+bi.
(5+2i)+(3+i)
(6+i)+(1+2i)
(4−3i)−(5+3i)
(3−5i)−(3+2i)
(−2−3i)+(−3−2i)
(−5−3i)+(2−i)
3(5+2i)
4(3+5i)
−4(2−3i)
−7(3−4i)
3i(5+i)
2i(4+3i)
4i(2−5i)
−3i(5−2i)
(3+i)(2+3i)
(4+3i)(2+5i)
(2−3i)(2+3i)
(4−3i)(4+3i)
(3+4i)(4−3i)
(−2+3i)(−3+10i)
(√3−12i)2
(−√5−13i)2
(2−√−16)(3+5i)
(5−2i)(3+√−25)
In Exercises 33–38, write the conjugate ˉz of each complex number z Then find zˉz.
z=2−3i
z=4+5i
z=12−2i
z=23+12i
z=√2−3i
z=√5+√3 i
In Exercises 39–52, write each quotient in the standard form a+bi.
5−i
2−3i
−11+i
12−i
5i2+i
3i2−i
2+3i1+i
3+5i4+i
2−5i4−7i
3+5i1−3i
2+√−41+i
5−√−93+2i
−2+√−252−3i
−5−√−45−√−9
In Exercises 53–62, solve each equation.
x2+5=1
4x2+9=0
z2−2z+2=0
x2−6x+11=0
2x2−20x+49=−7
4y2+4y+5=0
8(x2−x)=x2−3
t(t+1)=3t2+1
9k2+25=0
3k2+4=0
Series circuits. Assuming that the impedance of a resistor in a circuit is Z1=4+3i ohms and the impedance of a second resistor is Z2=5−2i ohms, find the total impedance of the two resistors when placed in series (sum of the two impedances).
Parallel circuits. If the two resistors in Exercise 63 are connected in parallel, the total impedance is given by
Find the total impedance, assuming that the resistors in Exercise 63 are connected in parallel.
As with impedance, the current I and voltage V in a circuit can be represented by complex numbers. The three quantities (voltage, V; impedance, Z; and current, I) are related by the equation Z=VI. Thus, if two of these values are given, the value of the third can be found from the equation.
In Exercises 65–70, use the equation V=ZI to find the value that is not specified.
Finding impedance: I=7+5iV=35+70i
Finding impedance: I=7+4iV=45+88i
Finding voltage: Z=5−7iI=2+5i
Finding voltage: Z=7−8iI=13+16i
Finding current: V=12+10iZ=12+6i
Finding current: V=29+18iZ=25+6i
In Exercises 71–80, find each power of i and simplify the expression.
i17
i125
i−7
i−24
i10+7
9+i3
3i5−2i3
5i6−3i4
2i3(1+i4)
5i5(i3−i)
Prove that the reciprocal of a+bi, where both a and b are nonzero, is aa2+b2−ba2+b2i.
In Exercises 82–84, let z=a+bi. Prove each statement.
a=z+ˉz2
b=z−ˉz2i
The product zˉz=0 if and only if z=0.
Show that 1−2i5−5i=2−3i9−7i.
Write in standard form:
Solve for z:
Let z=2−3i and w=1+2i.
Show that ¯(zw)=(ˉz)(ˉw).
Show that (zw)=ˉzˉw.
State whether each of the following is true or false. Explain your reasoning.
Every real number is a complex number.
Every complex number is a real number.
Every complex number is an imaginary number.
A real number is a complex number whose imaginary part is 0.
The product of a complex number and its conjugate is a real number.
The equality z=ˉz holds if and only if z is a real number.
Find the least positive n for which (1+i1−i)n=1.
Show that the set of complex numbers does not have the ordering properties of the set of real numbers. [Hint: Assume that ordering properties hold. Then by the law of trichotomy, i=0 or i<0 or i>0. Show that each leads to a contradiction.]
Suppose that multiplication of complex numbers had been defined as
Find two complex numbers z and w, z≠0,w≠0 but zw=0.
In Exercises 93 and 94, factor each expression by grouping.
x3−x2−9x+9
−x3−2x2+25x+50
In Exercises 95–100, find the LCD for each group of rational expressions.
1x−1, 1x
32−x, xx−2
15x+3, 2x+1x−3
3xx2−25, 95−x
x+1(x−3)(2−x), 12(x−2)(x+1)
x4x2−1, 72x−1, 1x
In Exercises 101–108, simplify each expression.
272/3
84/3
45/2
253/2
(√3+x2)2
(1+√x+1)2
((3x2+7)1/3)3
((5x−2)3/5)5/3