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No credit card required 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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