Download Complex Numbers - Stewart Calculus

COMPLEX NUMBERS
Im
2+3i
_4+2i
i
0
_i
_2-2i
Re
1
3-2i
FIGURE 1
Complex numbers as points in
the Argand plane
A complex number can be represented by an expression of the form a bi, where a and
b are real numbers and i is a symbol with the property that i 2 1. The complex number a bi can also be represented by the ordered pair a, b and plotted as a point in a
plane (called the Argand plane) as in Figure 1. Thus, the complex number i 0 1 i is
identified with the point 0, 1.
The real part of the complex number a bi is the real number a and the imaginary
part is the real number b. Thus, the real part of 4 3i is 4 and the imaginary part is 3.
Two complex numbers a bi and c di are equal if a c and b d, that is, their real
parts are equal and their imaginary parts are equal. In the Argand plane the horizontal axis
is called the real axis and the vertical axis is called the imaginary axis.
The sum and difference of two complex numbers are defined by adding or subtracting
their real parts and their imaginary parts:
a bi c di a c b di
a bi c di a c b di
For instance,
1 i 4 7i 1 4 1 7i 5 6i
The product of complex numbers is defined so that the usual commutative and distributive
laws hold:
a bic di ac di bic di
ac adi bci bdi 2
Since i 2 1, this becomes
a bic di ac bd ad bci
EXAMPLE 1
1 3i2 5i 12 5i 3i2 5i
2 5i 6i 151 13 11i
Division of complex numbers is much like rationalizing the denominator of a rational
expression. For the complex number z a bi, we define its complex conjugate to be
z a bi. To find the quotient of two complex numbers we multiply numerator and
denominator by the complex conjugate of the denominator.
EXAMPLE 2 Express the number
1 3i
in the form a bi.
2 5i
SOLUTION We multiply numerator and denominator by the complex conjugate of 2 5i,
namely 2 5i, and we take advantage of the result of Example 1:
Im
1 3i
1 3i 2 5i
13 11i
13
11
2
i
2 2 5i
2 5i
2 5i
2 5
29
29
z=a+bi
i
0
Re
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_i
The geometric interpretation of the complex conjugate is shown in Figure 2: z is the
reflection of z in the real axis. We list some of the properties of the complex conjugate in
the following box. The proofs follow from the definition and are requested in Exercise 18.
Properties of Conjugates
z=a-bi
–
zwzw
zw z w
zn zn
FIGURE 2
1
2 ■ COMPLEX NUMBERS
The modulus, or absolute value, z of a complex number z a bi is its distance
from the origin. From Figure 3 we see that if z a bi, then
Im
@ z=a+bi
„b„
„
+
„@
œ „a
bi
=
|z|
z sa
b
0
Re
a
2
b2
Notice that
zz a bia bi a 2 abi abi b 2i 2 a 2 b 2
FIGURE 3
zz z
and so
2
This explains why the division procedure in Example 2 works in general:
z
w
zw
ww
zw
w
2
Since i 2 1, we can think of i as a square root of 1. But notice that we also have
i2 i 2 1 and so i is also a square root of 1. We say that i is the principal
square root of 1 and write s1 i. In general, if c is any positive number, we write
sc sc i
With this convention, the usual derivation and formula for the roots of the quadratic equation ax 2 bx c 0 are valid even when b 2 4ac 0:
x
b sb 2 4ac
2a
EXAMPLE 3 Find the roots of the equation x 2 x 1 0.
SOLUTION Using the quadratic formula, we have
x
1 s1 2 4 1
1 s3
1 s3 i
2
2
2
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 bx 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 a n1 x n1 a 1 x 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.
POLAR FORM
We know that any complex number z a bi can be considered as a point a, b and that
any such point can be represented by polar coordinates r, with r 0. In fact,
Im
a+bi
r
a r cos Thomson Brooks-Cole copyright 2007
b
b r sin ¨
0
FIGURE 4
a
Re
as in Figure 4. Therefore, we have
z a bi r cos r sin i
COMPLEX NUMBERS ■ 3
Thus, we can write any complex number z in the form
z rcos i sin r z sa 2 b 2
where
tan and
b
a
The angle is called the argument of z and we write argz. Note that argz 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 i
SOLUTION
(b) w s3 i
(a) We have r z s12 12 s2 and tan 1, so we can take 4.
Therefore, the polar form is
Im
1+i
z s2
œ„
2
π
4
0
_
π
6
cos
i sin
4
4
(b) Here we have r w s3 1 2 and tan 1
s3 . Since w lies in the
fourth quadrant, we take 6 and
Re
2
w 2 cos œ„
3-i
6
i sin 6
The numbers z and w are shown in Figure 5.
FIGURE 5
The polar form of complex numbers gives insight into multiplication and division. Let
z1 r1cos 1 i sin 1 z2 r2cos 2 i sin 2 be two complex numbers written in polar form. Then
Im
z™
z1 z2 r1r2cos 1 i sin 1 cos 2 i sin 2 z¡
r1r2 cos 1 cos 2 sin 1 sin 2 isin 1 cos 2 cos 1 sin 2 ¨™
Therefore, using the addition formulas for cosine and sine, we have
¨¡
Re
¨¡+¨™
z1z2 r1r2 cos1 2 i sin1 2 1
z¡z™
This formula says that to multiply two complex numbers we multiply the moduli and add
the arguments. (See Figure 6.)
A similar argument using the subtraction formulas for sine and cosine shows that to
divide two complex numbers we divide the moduli and subtract the arguments.
FIGURE 6
Im
z
z1
r1
cos1 2 i sin1 2 z2
r2
r
¨
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0
_¨
1
r
Re
In particular, taking z1 1 and z2 z, (and therefore 1 0 and 2 ), we have the following, which is illustrated in Figure 7.
1
z
If
FIGURE 7
z2 0
z rcos i sin ,
then
1
1
cos i sin .
z
r
4 ■ COMPLEX NUMBERS
EXAMPLE 5 Find the product of the complex numbers 1 i and s3 i in polar form.
SOLUTION From Example 4 we have
i sin
4
4
1 i s2 cos
s3 i 2 cos and
Im
z=1+i
œ„
2
zw
2œ„2
So, by Equation 1,
π
12
0
1 i(s3 i) 2s2 cos
Re
2
2s2 cos
w=œ„
3-i
FIGURE 8
6
i sin 4
6
6
i sin
4
6
i sin
12
12
This is illustrated in Figure 8.
Repeated use of Formula 1 shows how to compute powers of a complex number. If
z r cos i sin then
z 2 r 2cos 2 i sin 2 and
z 3 zz 2 r 3cos 3 i sin 3 In general, we obtain the following result, which is named after the French mathematician
Abraham De Moivre (1667–1754).
2
De Moivre’s Theorem If z r cos i sin and n is a positive integer, then
z n r cos i sin n r ncos n i sin n This says that to take the nth power of a complex number we take the nth power of the
modulus and multiply the argument by n.
EXAMPLE 6 Find
SOLUTION Since
1
2
( 12 12 i)10.
12 i 12 1 i, it follows from Example 4(a) that 12 12 i has the polar
form
1
1
s2
i
2
2
2
cos
i sin
4
4
So by De Moivre’s Theorem,
1
1
i
2
2
10
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25
2 10
10
s2
2
cos
cos
10
10
i sin
4
4
5
5
i sin
2
2
1
i
32
De Moivre’s Theorem can also be used to find the n th roots of complex numbers. An n
th root of the complex number z is a complex number w such that
wn z
COMPLEX NUMBERS ■ 5
Writing these two numbers in trigonometric form as
w scos i sin and
z r cos i sin and using De Moivre’s Theorem, we get
s ncos n i sin n r cos i sin The equality of these two complex numbers shows that
sn r
and
cos n cos s r 1
n
or
sin n sin and
From the fact that sine and cosine have period 2 it follows that
n 2k
Thus
w r 1
n cos
2k
n
2k
n
or
2k
n
i sin
Since this expression gives a different value of w for k 0, 1, 2, . . . , n 1, we have the
following.
3 Roots of a Complex Number Let z r cos i sin and let n be a positive integer. Then z has the n distinct n th roots
wk r 1
n cos
2k
n
i sin
2k
n
where k 0, 1, 2, . . . , n 1.
Notice that each of the nth roots of z has modulus wk r 1
n. Thus, all the nth roots
of z lie on the circle of radius r 1
n in the complex plane. Also, since the argument of each
successive nth root exceeds the argument of the previous root by 2
n , we see that the n
th roots of z are equally spaced on this circle.
EXAMPLE 7 Find the six sixth roots of z 8 and graph these roots in the complex
plane.
SOLUTION In trigonometric form, z 8cos i sin . Applying Equation 3 with n 6,
we get
wk 8 1
6 cos
2k
2k
i sin
6
6
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We get the six sixth roots of 8 by taking k 0, 1, 2, 3, 4, 5 in this formula:
w0 8 1
6 cos
i sin
6
6
w1 8 1
6 cos
i sin
2
2
w2 8 1
6 cos
5
5
i sin
6
6
s2
1
s3
i
2
2
s2 i
s2 1
s3
i
2
2
6 ■ COMPLEX NUMBERS
Im
œ„2 i w¡
w™
2
_œ„
w¸
0
w∞
_œ„2 i
7
7
i sin
6
6
w4 8 1
6 cos
3
3
i sin
2
2
w5 8 1
6 cos
11
11
i sin
6
6
œ„
2 Re
w£
w¢
w3 8 1
6 cos
s2 1
s3
i
2
2
s2 i
s2
1
s3
i
2
2
All these points lie on the circle of radius s2 as shown in Figure 9.
FIGURE 9
The six sixth roots of z=_8
COMPLEX EXPONENTIALS
We also need to give a meaning to the expression e z when z x iy is a complex number. The theory of infinite series as developed in Chapter 8 can be extended to the case
2 as our guide,
where the terms are complex numbers. Using the Taylor series for e x (8.7.12)
we define
ez 4
n0
zn
z2
z3
1z
n!
2!
3!
and it turns out that this complex exponential function has the same properties as the real
exponential function. In particular, it is true that
e z z e z e z
5
1
2
1
2
If we put z iy, where y is a real number, in Equation 4, and use the facts that
i 2 1,
i 3 i 2i i,
i 4 1, i 5 i, . . .
we get
e iy 1 iy 1 iy 1
iy2
iy3
iy4
iy5
2!
3!
4!
5!
y2
y3
y4
y5
i
i
2!
3!
4!
5!
y2
y4
y6
y3
y5
i y 2!
4!
6!
3!
5!
cos y i sin y
Here we have used the Taylor series for cos y and sin y (Equations 8.7.17 and 8.7.16).
The result is a famous formula called Euler’s formula:
6
e iy cos y i sin y
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Combining Euler’s formula with Equation 5, we get
7
e xiy e xe iy e x cos y i sin y
COMPLEX NUMBERS ■ 7
(a) e i
EXAMPLE 8 Evaluate:
■ ■
(b) e1i
2
SOLUTION
We could write the result of Example 8(a)
as
(a) From Euler’s equation (6) we have
e i 1 0
e i cos i sin 1 i0 1
This equation relates the five most famous
numbers in all of mathematics: 0, 1, e, i, and .
(b) Using Equation 7 we get
i sin
2
2
e1i
2 e1 cos
1
i
0 i1 e
e
Finally, we note that Euler’s equation provides us with an easier method of proving
De Moivre’s Theorem:
r cos i sin n re i n r ne in r ncos n i sin n EXERCISES
A Click here for answers.
25–28
Click here for solutions.
S
5
3. 2 5i 4 i
4. 1 2i 8 3i 5. 12 7i
6. 2i ( 12 i )
1 4i
7.
3 2i
3 2i
8.
1 4i
9.
27. 3 4i
2. (4 2 i) (9 2 i)
1
1
1i
■
■
11. i 3
12. i 100
13. s25
14. s3 s12
■
■
15–17
■
■
■
■
■
■
■
■
■
■
32. z 4(s3 i ),
w 3 3i
■
■
■
■
■
■
■
■
■
■
■
■
■
■
■
■
21. x 2x 5 0
22. 2x 2x 1 0
23. z z 2 0
24. z z 0
2
■
1
2
■
■
■
■
■
■
■
■
■
■
■
■
■
■
■
■
■
5
■
■
■
■
■
Find the indicated roots. Sketch the roots in the complex
38. The fifth roots of 32
■
■
■
■
40. The cube roots of 1 i
■
■
■
43. e i
3
44. e i
45. e 2i
46. e i
■
■
■
■
■
■
■
■
■
■
Write the number in the form a bi.
42. e 2 i
■
■
■
■
■
■
■
■
47. Use De Moivre’s Theorem with n 3 to express cos 3 and
1
4
■
■
36. 1 i 8
5
i
2
41. e
20. x 1
■
■
41–46
19. 4x 9 0
2
■
34. (1 s3 i )
33. 1 i 20
■
4
2
■
Find the indicated power using De Moivre’s Theorem.
39. The cube roots of i
Find all solutions of the equation.
2
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■
■
(c) z z , where n is a positive integer
[Hint: Write z a bi, w c di.]
2
■
plane.
n
n
■
■
37. The eighth roots of 1
■
■
w 1 s3 i
w 1 i
37–40
16. 1 2 s2 i
18. Prove the following properties of complex numbers.
(a) z w z w
(b) zw z w
■
■
31. z 2 s3 2i,
■
17. 4i
19–24
■
w 8i
35. (2 s3 2i )
number.
■
■
30. z 4 s3 4i,
33–36
Find the complex conjugate and the modulus of the
15. 12 5i
■
29. z s3 i,
■
■
28. 8i
■
29–32
Find polar forms for z w, z
w, and 1
z by first putting z
and w into polar form.
3
4 3i
10.
26. 1 s3 i
25. 3 3i
1–14
Evaluate the expression and write your answer in the
form a bi.
1. 5 6i 3 2i Write the number in polar form with argument between 0
and 2.
■
■
sin 3 in terms of cos and sin .
8 ■ COMPLEX NUMBERS
48. Use Euler’s formula to prove the following formulas for cos x
and sin x :
cos x e ix eix
2
sin x e ix eix
2i
49. If ux f x itx is a complex-valued function of a real
Thomson Brooks-Cole copyright 2007
variable x and the real and imaginary parts f x and tx are
differentiable functions of x, then the derivative of u is defined
to be ux f x itx. Use this together with Equation 7
to prove that if Fx e rx, then Fx re rx when r a bi
is a complex number.
50. (a) If u is a complex-valued function of a real variable, its
indefinite integral x ux dx is an antiderivative of u.
Evaluate
ye
1i x
dx
(b) By considering the real and imaginary parts of the integral
in part (a), evaluate the real integrals
ye
x
cos x dx
and
ye
x
sin x dx
(c) Compare with the method used in Example 4 in Section 6.1.
COMPLEX NUMBERS ■ 9
ANSWERS
S
1. 8 4i
1
2
(2 s2 )cos13
12 i sin13
12, 14 cos
6 i sin
6
7.
12 i
11
13
10
13
11. i
37. 1, i, (1
s2 )1 i i
13. 5i
15. 12 5i; 13
3
23. 2 (s7
2)i
21. 1 2i
39. (s3
2) 2 i, i
1
Im
Im
i
19. 2 i
17. 4i, 4
35. 512 s3 512i
33. 1024
3. 13 18i
5. 12 7i
9.
31. 4 s2 cos7
12 i sin7
12,
Click here for solutions.
1
0
1
0
Re
25. 3 s2 cos3
4 i sin3
4
[
27. 5{cos tan
1 4
3
29. 4cos
2 i sin
2, cos
6 i sin
6,
Thomson Brooks-Cole copyright 2007
1
2
_i
( )] i sin[tan ( )]}
1 4
3
cos
6 i sin
6
Re
41. i
43.
1
2
(s3
2) i
45. e 2
47. cos 3 cos3 3 cos sin2, sin 3 3 cos2 sin sin3
10 ■ COMPLEX NUMBERS
SOLUTIONS
1. (5 − 6i) + (3 + 2i) = (5 + 3) + (−6 + 2)i = 8 + (−4)i = 8 − 4i
3. (2 + 5i)(4 − i) = 2(4) + 2(−i) + (5i)(4) + (5i)(−i) = 8 − 2i + 20i − 5i2
= 8 + 18i − 5(−1) = 8 + 18i + 5 = 13 + 18i
5. 12 + 7i = 12 − 7i
1 + 4i 3 − 2i
3 − 2i + 12i − 8(−1)
11
10
11 + 10i
1 + 4i
=
·
=
=
+ i
=
3 + 2i
3 + 2i 3 − 2i
32 + 22
13
13
13
1
1−i
1−i
1−i
1
1
1
=
·
=
=
= − i
9.
1+i
1+i 1−i
1 − (−1)
2
2
2
7.
11. i3 = i2 · i = (−1)i = −i
√
√
13. −25 = 25 i = 5i
s
√
√
122 + (−5)2 = 144 + 25 = 169 = 13
t
√
17. −4i = 0 − 4i = 0 + 4i = 4i and |−4i| = 02 + (−4)2 = 16 = 4
t
t
19. 4x2 + 9 = 0 ⇔ 4x2 = −9 ⇔ x2 = − 94 ⇔ x = ± − 94 = ± 94 i = ± 32 i.
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15. 12 − 5i = 12 + 15i and |12 − 15i| =
21. By the quadratic formula, x2 + 2x + 5 = 0 ⇔
s
√
−2 ± 22 − 4(1)(5)
−2 ± −16
−2 ± 4i
x=
=
=
= −1 ± 2i.
2(1)
2
2
s
√
√
−1 ± 12 − 4(1)(2)
−1 ± −7
1
7
2
23. By the quadratic formula, z + z + 2 = 0 ⇔ z =
=
=− ±
i.
2(1)
2
2
2
s
√
3
25. For z = −3 + 3i, r = (−3)2 + 32 = 3 2 and tan θ = −3
= −1 ⇒ θ = 3π
4 (since z lies in the second
√ 3π
3π
quadrant). Therefore, −3 + 3i = 3 2 cos 4 + i sin 4 .
√
27. For z = 3 + 4i, r = 32 + 42 = 5 and tan θ = 43 ⇒ θ = tan−1 43 (since z lies in the first quadrant).
Therefore, 3 + 4i = 5 cos tan−1 43 + i sin tan−1 43 .
t√ √
2
29. For z = 3 + i, r =
3 + 12 = 2 and tan θ = √13 ⇒ θ = π6 ⇒ z = 2 cos π6 + i sin π6 .
√
√
For w = 1 + 3 i, r = 2 and tan θ = 3 ⇒ θ = π3 ⇒ w = 2 cos π3 + i sin π3 . Therefore,
zw = 2 · 2 cos π6 + π3 + i sin π6 + π3 = 4 cos π2 + i sin π2 ,
z/w = 22 cos π6 − π3 + i sin π6 − π3 = cos − π6 + i sin − π6 , and 1 = 1 + 0i = 1(cos 0 + i sin 0) ⇒
1/z = 12 cos 0 − π6 + i sin 0 − π6 = 12 cos − π6 + i sin − π6 . For 1/z, we could also use the formula that
precedes Example 5 to obtain 1/z = 12 cos π6 − i sin π6 .
t √ √
2
√ = − √1
2 3 + (−2)2 = 4 and tan θ = 2−2
31. For z = 2 3 − 2i, r =
3
3
√
⇒ θ = − π6 ⇒ z = 4 cos − π6 + i sin − π6 . For w = −1 + i, r = 2,
√ 1
3π
tan θ = −1
. Therefore,
= −1 ⇒ θ = 3π
⇒ w = 2 cos 3π
4
4 + i sin 4
√ π
√ π
3π
3π
7π
zw = 4 2 cos − 6 + 4 + i sin − 6 + 4
= 4 2 cos 12 + i sin 7π
,
12
π
π
11π 11π 4
3π
3π
4
z/w = √2 cos − 6 − 4 + i sin − 6 − 4
= √2 cos − 12 + i sin − 12
√ 13π
13π
= 2 2 cos 12 + i sin 12 , and
1/z = 14 cos − π6 − i sin − π6 = 14 cos π6 + i sin π6 .
√ √
33. For z = 1 + i, r = 2 and tan θ = 11 = 1 ⇒ θ = π4 ⇒ z = 2 cos π4 + i sin π4 . So by
De Moivre’s Theorem,
k√ l20 1/2 20 cos 204· π + i sin 204· π
2 cos π4 + i sin π4
= 2
(1 + i)20 =
= 210 (cos 5π + i sin 5π) = 210 [−1 + i(0)] = −210 = −1024
COMPLEX NUMBERS ■ 11
t √ √
√
2
2
2 3 + 22 = 16 = 4 and tan θ = 2√
35. For z = 2 3 + 2i, r =
= √13 ⇒ θ = π6 ⇒
3
z = 4 cos π6 + i sin π6 . So by De Moivre’s Theorem,
√
k √
5 l
√
5
2 3 + 2i = 4 cos π6 + i sin π6
= 1024 − 23 + 12 i = −512 3 + 512i
= 45 cos 5π
+ i sin 5π
6
6
37. 1 = 1 + 0i = 1 (cos 0 + i sin 0). Using Equation 3 with r = 1, n = 8, and θ = 0, we have
kπ
kπ
0 + 2kπ
0 + 2kπ
1/8
cos
wk = 1
+ i sin
= cos
+ i sin
, where k = 0, 1, 2, . . . , 7.
8
8
4
4
w0 = 1(cos 0 + i sin 0) = 1, w1 = 1 cos π4 + i sin π4 = √12 + √12 i,
= − √12 + √12 i,
w2 = 1 cos π2 + i sin π2 = i, w3 = 1 cos 3π
+ i sin 3π
4
4
5π
w4 = 1(cos π + i sin π) = −1, w5 = 1 cos 5π
= − √12 − √12 i,
4 + i sin 4
= −i, w7 = 1 cos 7π
= √12 − √12 i
w6 = 1 cos 3π
+ i sin 3π
+ i sin 7π
2
2
4
4
39. i = 0 + i = 1 cos π2 + i sin π2 . Using Equation 3 with r = 1, n = 3, and θ =
π
π
+ 2kπ
+ 2kπ
wk = 11/3 cos 2
+ i sin 2
, where k = 0, 1, 2.
3
3
√
w0 = cos π6 + i sin π6 = 23 + 12 i
√
5π
= − 23 + 12 i
w1 = cos 5π
6 + i sin 6
9π
= −i
w2 = cos 9π
6 + i sin 6
41. Using Euler’s formula (6) with y =
π
,
2
we have
π
,
2
we have eiπ/2 = cos π2 + i sin π2 = 0 + 1i = i.
√
π
π
1
π
3
43. Using Euler’s formula (6) with y = , we have eiπ/3 = cos + i sin = +
i.
3
3
3
2
2
45. Using Equation 7 with x = 2 and y = π, we have e2+iπ = e2 eiπ = e2 (cos π + i sin π) = e2 (−1 + 0) = −e2 .
47. Take r = 1 and n = 3 in De Moivre’s Theorem to get
[1(cos θ + i sin θ)]3 = 13 (cos 3θ + i sin 3θ)
(cos θ + i sin θ)3 = cos 3θ + i sin 3θ
cos3 θ + 3 cos2 θ (i sin θ) + 3(cos θ)(i sin θ)2 + (i sin θ)3 = cos 3θ + i sin 3θ
cos3 θ + 3 cos2 θ sin θ i − 3 cos θ sin2 θ − sin3 θ i = cos 3θ + i sin 3θ
3
cos θ − 3 sin2 θ cos θ + 3 sin θ cos2 θ − sin3 θ i = cos 3θ + i sin 3θ
Equating real and imaginary parts gives cos 3θ = cos3 θ − 3 sin2 θ cos θ
and sin 3θ = 3 sin θ cos2 θ − sin3 θ
49. F (x) = erx = e(a+bi)x = eax+bxi = eax (cos bx + i sin bx) = eax cos bx + i(eax sin bx) ⇒
F 0 (x) = (eax cos bx)0 + i(eax sin bx)0
= (aeax cos bx − beax sin bx) + i(aeax sin bx + beax cos bx)
= a [eax (cos bx + i sin bx)] + b [eax (− sin bx + i cos bx)]
= aerx + b eax i2 sin bx + i cos bx
Thomson Brooks-Cole copyright 2007
= aerx + bi [eax (cos bx + i sin bx)] = aerx + bierx = (a + bi)erx = rerx