James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

appendix h Complex Numbers A59
We observe that the solutions of the equation in Example 3 are complex conjugates of
each other. In general, the solutions of any quadratic equation ax 2 1 bx 1 c − 0 with
real coefficients a, b, and c are always complex conjugates. (If z is real, z − z, so z is its
own conjugate.)
We have seen that if we allow complex numbers as solutions, then every quadratic
equation has a solution. More generally, it is true that every polynomial equation
a n x n 1 a n21 x n21 1 ∙ ∙ ∙ 1 a 1 x 1 a 0 − 0
of degree at least one has a solution among the complex numbers. This fact is known as
the Fundamental Theorem of Algebra and was proved by Gauss.
Im
r
¨
0 a
a+bi
b
Re
Polar Form
We know that any complex number z − a 1 bi can be considered as a point sa, bd and
that any such point can be represented by polar coordinates sr, d with r > 0. In fact,
as in Figure 4. Therefore we have
a − r cos
b − r sin
FIGURE 4
z − a 1 bi − sr cos d 1 sr sin di
Thus we can write any complex number z in the form
z − rscos 1 i sin d
where r − | z | − sa 2 1 b 2 and tan − b a
The angle is called the argument of z and we write − argszd. Note that argszd is not
unique; any two arguments of z differ by an integer multiple of 2.
EXAMPLE 4 Write the following numbers in polar form.
(a) z − 1 1 i
(b) w − s3 2 i
SOLUTION
Im
0
œ„2
π
4
_ π 6
2
1+i
Re
œ„3-i
(a) We have r − | z | − s12 1 1 2 − s2 and tan − 1, so we can take − y4.
Therefore the polar form is
z − s2 Scos 4 1 i sin 4D
(b) Here we have r − | w | − s3 1 1 − 2 and tan − 21ys3 . Since w lies in the
fourth quadrant, we take − 2y6 and
w − 2FcosS2 6D1i sinS2 6DG
FIGURE 5
The numbers z and w are shown in Figure 5.
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