44 CHAPTER 5

The situation with cubic equations was more complicated.

Mathematicians discovered a formula for solving cubic equations

similar to but more complicated than the quadratic formula

−b±

√

b

2

−4ac

2a

which solves the quadratic equation ax

2

+ bx + c = 0. In

some cases, however, some of the numbers in the cubic formula

were square roots of negative numbers, even though in the end all

of the solutions to the cubic equation were real numbers. We are not

going to go through the algebra, but here is a very explicit example.

The equation x

3

− 7x + 6 = 0 has the three solutions x = 1, x = 2,

and x =−3. If you try to solve this equation by using the cubic

formula (which is like the quadratic formula, only much more com-

plicated), along the way you unavoidably encounter the square root

of a negative number.

Complex Arithmetic

Whether or not we believe that complex numbers really exist, we

need to come up with rules for manipulating them. And the rules

are simple enough.

DEFINITION: A complex number is a number of the form

a + bi,wherea and b are real numbers.

So 3 + 4i and

√

2 + πi are complex numbers;

3

so is 0 + 0i (which

is another w ay of writing 0). Sometimes, we write the “i” before the

second number: 2 − i sin 2 is also a complex number.

4

In the complex number a + bi, we call a the real part and bi

the imaginary part. (Actually, mathematicians normally reserve

the term “imaginary part” for the real number b, but that is a bit

confusing, and we will not follow that usage.) We consider every

real number x to be a complex number too, but write x instead of

x + 0i.

3

The symbol π stands for the area of a circle of radius 1, which is approximately

3.14159265. It is a nonrepeating inﬁnite decimal.

4

If b is negative, we can write a − (−b)i rather than a + bi.

COMPLEX NUMBERS 45

Now we need to know how to do arithmetic with complex

numbers. Addition is simple: we just add the real and the imagi-

nary parts separately. In other words, (a + bi) + (c + di) = (a + c) +

(b + d)i.

EXERCISE: What is the sum of 2 + 3i and 4 − 7i?

SOLUTION: We add each part, and get

(2 + 3i) + (4 − 7i) = (2 + 4) + (3 − 7)i = 6 − 4i.

Subtraction is done similarly; each piece is subtracted separately.

Formally, we write (a + bi) − (c + di) = (a − c) + (b − d)i.

EXERCISE: Subtract 4 − 7i from 2 + 3i.

SOLUTION: We subtract each part, and get

(2 + 3i) − (4 − 7i) = (2 − 4) + (3 − (−7))i =−2 + 10i.

Multiplication is trickier. W e use the distributive la w of multipli-

cation over addition (that is the one that says A(B + C) = AB + AC)

and whenever we get i · i (which is also written i

2

), we replace it

by −1. In other words, we adopt the convention that i

2

=−1.

Formally, the rule looks like

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

because we get the term (bi)(di) = bdi

2

=−bd.

EXERCISE: What is the product of 2 + 3i and 4 − 7i?

SOLUTION: Use the formula to get

(2 + 3i)(4 − 7i) = (2 · 4 − 3 · (−7)) + (2 · (−7) + 3 · 4)i = 29 − 2i.

Finally, there is division. Rather than tell how to divide one

complex number by another, we give a formula for the reciprocal

of a complex number. Then, just like with division of fractions,

rather than dividing by a complex number, you can multiply by the

reciprocal.

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