This document discusses polar form and DeMoivre's theorem for complex numbers. It begins by introducing polar form, which represents a complex number z as z = r(cosθ + i sinθ) where r is the modulus and θ is the argument. It then states DeMoivre's theorem, which says that for any positive integer n, zn = rn(cosnθ + i sinnθ). An example calculates (−1 + √3i)12 by first converting to polar form and then applying DeMoivre's theorem.

1. SECTION 8.3 POLAR FORM AND DEMOIVRE’S THEOREM 445
1ϩi
΂ ͙2 ΃
32. Describe the set of points in the complex plane that satisfy the 2
following. 34. (a) Verify that ϭ i.
(a) ͉z͉ ϭ 4 (b) ͉z Ϫ i͉ ϭ 2 (b) Find the two square roots of i.
(c) ͉z ϩ 1͉Յ 1 (d) ͉z͉ Ͼ 3 (c) Find all zeros of the polynomial x 4 ϩ 1.
33. (a) Evaluate ͑1͞i͒ for n ϭ 1, 2, 3, 4, and 5.
n
(b) Calculate ͑1͞i͒57 and ͑1͞i͒1995.
(c) Find a general formula for ͑1͞i͒n for any positive integer n.
8.3 POLAR FORM AND DEMOIVRE’S THEOREM
Figure 8.6 At this point we can add, subtract, multiply, and divide complex numbers. However, there
Imaginary
axis is still one basic procedure that is missing from our algebra of complex numbers. To see
this, consider the problem of finding the square root of a complex number such as i. When
(a, b)
we use the four basic operations (addition, subtraction, multiplication, and division), there
seems to be no reason to guess that
1ϩi 1ϩi
΂ ͙2 ΃
r 2
b ͙i ϭ . That is, ϭ i.
͙2
θ
Real To work effectively with powers and roots of complex numbers, it is helpful to use a polar
a 0 axis representation for complex numbers, as shown in Figure 8.6. Specifically, if a ϩ bi is a
nonzero complex number, then we let ␪ be the angle from the positive x-axis to the radial
Complex Number: a + bi line passing through the point (a, b) and we let r be the modulus of a ϩ bi. Thus,
Rectangular Form: (a, b)
Polar Form: (r, θ )
a ϭ r cos ␪, b ϭ r sin ␪, and r ϭ ͙a 2 ϩ b2
and we have a ϩ bi ϭ ͑r cos ␪͒ ϩ ͑r sin ␪͒i from which we obtain the following polar
form of a complex number.
Definition of Polar Form The polar form of the nonzero complex number z ϭ a ϩ bi is given by
of a Complex Number z ϭ r ͑cos ␪ ϩ i sin ␪͒
where a ϭ r cos ␪, b ϭ r sin ␪, r ϭ ͙a 2 ϩ b2, and tan␪ ϭ b͞a. The number r is the
modulus of z and ␪ is called the argument of z.

2. 446 CHAPTER 8 COMPLEX VECTOR SPACES
REMARK: The polar form of z ϭ 0 is given by z ϭ 0͑cos ␪ ϩ i sin ␪͒ where ␪ is any
angle.
Because there are infinitely many choices for the argument, the polar form of a complex
number is not unique. Normally, we use values of ␪ that lie between Ϫ␲ and ␲, though on
occasion it is convenient to use other values. The value of ␪ that satisfies the inequality
Ϫ␲ < ␪ ≤ ␲ Principal argument
is called the principal argument and is denoted by Arg(z). Two nonzero complex numbers
in polar form are equal if and only if they have the same modulus and the same principal
argument.
EXAMPLE 1 Finding the Polar Form of a Complex Number
Find the polar form of the following complex numbers. (Use the principal argument.)
(a) 1 Ϫ i (b) 2 ϩ 3i (c) i
Solution (a) We have a ϭ 1 and b ϭ Ϫ1, so r 2 ϭ 1 2 ϩ ͑Ϫ1͒2 ϭ 2, which implies that r ϭ ͙2.
From a ϭ r cos ␪ and b ϭ r sin ␪, we have
a 1 ͙2 b 1 ͙2
cos ␪ ϭ ϭ ϭ and sin ␪ ϭ ϭϪ ϭϪ .
r ͙2 2 r ͙2 2
Thus, ␪ ϭ Ϫ␲͞4 and
␲ ␲
΄ ΂ ΃
z ϭ ͙2 cos Ϫ ϩ i sin Ϫ
4 4
. ΂ ΃΅
(b) Since a ϭ 2 and b ϭ 3, we have r 2 ϭ 2 2 ϩ 32 ϭ 13, which implies that r ϭ ͙13.
Therefore,
a 2 b 3
cos ␪ ϭ ϭ and sin ␪ ϭ ϭ
r ͙13 r ͙13
and it follows that ␪ ϭ arctan (3/2). Therefore, the polar form is
΄ ΂ ΃ ΂ ΃΅
3 3
z ϭ ͙13 cos arctan ϩ i sin arctan
2 2
ഠ͙13 ͓cos ͑0.98͒ ϩ i sin ͑0.98͔͒.
(c) Since a ϭ 0 and b ϭ 1, it follows that r ϭ 1 and ␪ ϭ ␲͞2, so we have
␲ ␲
΂
z ϭ 1 cos
2
ϩ i sin .
2 ΃
The polar forms derived in parts (a), (b), and (c) are depicted graphically in Figure 8.7.

3. SECTION 8.3 POLAR FORM AND DEMOIVRE’S THEOREM 447
Figure 8.7
Imaginary Imaginary Imaginary
axis axis axis
4
z = 2 + 3i z=i
3 1
Real 2
θ axis
−1 1
θ θ
z=1−i Real Real
−2 axis axis
1 2
(a) z = 2 cos − π + i sin − π
[ ( ) ( )] (b) z = 13[cos(0.98) (c) z = 1 cos π + i sin π
( )
4 4 2 2
+ i sin(0.98)]
EXAMPLE 2 Converting from Polar to Standard Form
Express the following complex number in standard form.
␲ ␲
΄ ΂ ΃
z ϭ 8 cos Ϫ ϩ i sin Ϫ
3 3 ΂ ΃΅
Solution Since cos(Ϫ␲͞3͒ ϭ 1͞2 and sin ͑Ϫ␲͞3͒ ϭ Ϫ͙3͞2, we obtain the standard form
΂ ΃΅ ϭ 8 [ 2 Ϫ i ]
␲ ␲ ͙3
΄ ΂ ΃
1
z ϭ 8 cos Ϫ ϩ i sin Ϫ ϭ 4 Ϫ 4͙3i.
3 3 2
The polar form adapts nicely to multiplication and division of complex numbers.
Suppose we are given two complex numbers in polar form
z1 ϭ r1 ͑cos␪1 ϩ i sin␪1͒ and z 2 ϭ r2 ͑cos␪2 ϩ i sin␪2͒.
Then the product of z1 and z 2 is given by
z1 z 2 ϭ r1 r2 ͑cos␪1 ϩ i sin␪1͒͑cos␪2 ϩ i sin␪2͒
ϭ r1r2 ͓͑cos ␪1 cos ␪2 Ϫ sin ␪1 sin ␪2͒ ϩ i ͑cos ␪1 sin ␪2 ϩ sin ␪1 cos ␪2͔͒.
Using the trigonometric identities
cos͑␪1 ϩ ␪2͒ ϭ cos ␪1 cos ␪2 Ϫ sin ␪1 sin ␪2
and
sin ͑␪1 ϩ ␪2͒ ϭ sin ␪1 cos ␪2 ϩ cos ␪1 sin ␪2
we have
z1z 2 ϭ r1r2 ͓cos͑␪1 ϩ ␪2͒ ϩ i sin͑␪1 ϩ ␪2͔͒.
This establishes the first part of the following theorem. The proof of the second part is left
to you. (See Exercise 63.)

4. 448 CHAPTER 8 COMPLEX VECTOR SPACES
Theorem 8.4 Given two complex numbers in polar form
Product and Quotient of z1 ϭ r1͑cos ␪1 ϩ i sin ␪1͒ and z2 ϭ r2 ͑cos ␪2 ϩ i sin ␪ 2͒
Two Complex Numbers the product and quotient of the numbers are as follows.
z1z2 ϭ r1r2 ͓cos͑␪1 ϩ ␪2͒ ϩ i sin͑␪1 ϩ ␪2͔͒ Product
z1 r1
z 2 ϭ r2 ͓cos͑␪1 Ϫ ␪2͒ ϩ i sin͑␪1 Ϫ ␪2͔͒, z2 0 Quotient
This theorem says that to multiply two complex numbers in polar form, we multiply
moduli and add arguments, and to divide two complex numbers, we divide moduli and sub-
tract arguments. (See Figure 8.8.)
Figure 8.8
Imaginary Imaginary
axis axis
z1z2 z2 z1 z2
θ1 + θ 2 r r2 r1 z1
2
z1 z2
r1r2 θ2 r1 r1 θ2 r2
θ1
θ1 θ1 − θ 2
Real Real
axis axis
To multiply z1 and z2: To divide z1 by z2:
Multiply moduli and add arguments. Divide moduli and add arguments.
EXAMPLE 3 Multiplying and Dividing in Polar Form
Determine z1z2 and z1͞z2 for the complex numbers
␲ ␲ ␲ ␲
΂ ΃ ΂ ΃
1
z1 ϭ 5 cos ϩ i sin and z2 ϭ cos ϩ i sin .
4 4 3 6 6
Solution Since we are given the polar forms of z 1 and z 2, we can apply Theorem 8.4 as follows.
multiply
}
␲ ␲ ␲ ␲ 5␲ 5␲
΂3΃ ΄cos΂ 4 ϩ 6 ΃ ϩ i sin΂ 4 ϩ 6 ΃΅ ϭ 3 ΂cos 12 ϩ i sin 12 ΃
1 5
z1z2 ϭ ͑5͒
}
}
add add
divide
}
␲ ␲ ␲ ␲ ␲ ␲
΄ ΂ ΃ ΂ ΃΅ ϭ 15΂cos 12 ϩ i sin 12΃
z1 5
ϭ cos Ϫ ϩ i sin Ϫ
z2 1͞3 4 6 4 6
}
}
subtract subtract

5. SECTION 8.3 POLAR FORM AND DEMOIVRE’S THEOREM 449
REMARK: Try performing the multiplication and division in Example 3 using the stan-
dard forms
5͙2 5͙2 ͙3 1
z1 ϭ ϩ i and z2 ϭ ϩ i.
2 2 6 6
DeMoivre’s Theorem
Our final topic in this section involves procedures for finding powers and roots of complex
numbers. Repeated use of multiplication in the polar form yields
z ϭ r ͑cos␪ ϩ i sin␪͒
z 2 ϭ r ͑cos␪ ϩ i sin␪͒ r ͑cos␪ ϩ i sin␪͒ ϭ r 2 ͑cos 2␪ ϩ i sin 2␪͒
z3 ϭ r ͑cos ␪ ϩ i sin ␪͒ r 2 ͑cos 2␪ ϩ i sin 2␪͒ ϭ r 3 ͑cos 3␪ ϩ i sin 3␪͒.
Similarly,
z4 ϭ r4 ͑cos 4␪ ϩ i sin 4␪͒
z 5 ϭ r 5͑cos 5␪ ϩ i sin 5␪͒.
This pattern leads to the following important theorem, named after the French mathemati-
cian Abraham DeMoivre (1667–1754). You are asked to prove this theorem in Chapter
Review Exercise 71.
Theorem 8.5 If z ϭ r ͑cos ␪ ϩ i sin ␪͒ and n is any positive integer, then
DeMoivre’s Theorem zn ϭ r n͑cos n␪ ϩ i sin n␪͒.
EXAMPLE 4 Raising a Complex Number to an Integer Power
Find ͑Ϫ1 ϩ ͙3i͒12 and write the result in standard form.
Solution We first convert to polar form. For Ϫ1 ϩ ͙3i, we have
͙3
r ϭ ͙ ͑Ϫ1͒2 ϩ ͙͑3 ͒2 ϭ 2 and tan␪ ϭ ϭ Ϫ͙3
Ϫ1
which implies that ␪ ϭ 2␲͞3. Therefore,
2␲ 2␲
Ϫ1 ϩ ͙3i ϭ 2 cos ΂ 3
ϩ i sin
3
. ΃
By DeMoivre’s Theorem, we have
2␲ 2␲
΄΂ ΃΅
12
͑Ϫ1 ϩ ͙3i͒12 ϭ 2 cos ϩ i sin
3 3
12(2␲) 12(2␲)
΄
ϭ 212 cos
3
ϩ i sin
3 ΅

6. 450 CHAPTER 8 COMPLEX VECTOR SPACES
ϭ 4096͑cos 8␲ ϩ i sin 8␲͒
ϭ 4096͓1 ϩ i ͑0)͔ ϭ 4096.
Recall that a consequence of the Fundamental Theorem of Algebra is that a polynomial
of degree n has n zeros in the complex number system. Hence, a polynomial like
p͑x͒ ϭ x6 Ϫ 1 has six zeros, and in this case we can find the six zeros by factoring and
using the quadratic formula.
x6 Ϫ 1 ϭ ͑x3 Ϫ 1͒͑x3 ϩ 1͒ ϭ ͑x Ϫ 1͒͑x 2 ϩ x ϩ 1͒͑x ϩ 1͒͑x 2 Ϫ x ϩ 1͒
Consequently, the zeros are
Ϫ1 Ϯ ͙3i 1 Ϯ͙3i
x ϭ Ϯ1, xϭ , and xϭ .
2 2
Each of these numbers is called a sixth root of 1. In general, we define the nth root of a
complex number as follows.
Definition of n th Root of The complex number w ϭ a ϩ bi is an nth root of the complex number z if
a Complex Number z ϭ w n ϭ ͑a ϩ bi͒n .
DeMoivre’s Theorem is useful in determining roots of complex numbers. To see how this
is done, let w be an nth root of z, where
w ϭ s͑cos ␤ ϩ i sin ␤͒ and z ϭ r͑cos ␪ ϩ i cos ␪͒.
Then, by DeMoivre’s Theorem we have w n ϭ s n ͑cos n␤ ϩ i sin n␤͒ and since w n ϭ z, it
follows that
s n ͑cos n␤ ϩ i sin n␤͒ ϭ r͑cos ␪ ϩ i sin ␪͒.
Now, since the right and left sides of this equation represent equal complex numbers, we
n
can equate moduli to obtain sn = r which implies that s ϭ ͙r and equate principal
arguments to conclude that ␪ and n␤ must differ by a multiple of 2␲. Note that r is a
n
positive real number and hence s ϭ ͙r is also a positive real number. Consequently, for
some integer k, n␤ ϭ ␪ ϩ 2␲k, which implies that
␪ ϩ 2␲k
␤ϭ .
n
Finally, substituting this value for ␤ into the polar form of w, we obtain the result stated in
the following theorem.

7. SECTION 8.3 POLAR FORM AND DEMOIVRE’S THEOREM 451
Theorem 8.6 For any positive integer n, the complex number
n th Roots of a Complex z ϭ r ͑cos ␪ ϩ i sin ␪͒
Number has exactly n distinct roots. These n roots are given by
␪ ϩ 2␲ k ␪ ϩ 2␲ k
΄ ΂ ΃ ΂ ΃΅
n
͙r cos ϩ i sin
n n
where k ϭ 0, 1, 2, . . . , n Ϫ 1.
Figure 8.9 REMARK: Note that when k exceeds n Ϫ 1, the roots begin to repeat. For instance, if
Imaginary k ϭ n, the angle is
axis
␪ ϩ 2␲ n ␪
ϭ ϩ 2␲
n n
2π
n n which yields the same value for the sine and cosine as k ϭ 0.
r 2π
n Real
axis The formula for the nth roots of a complex number has a nice geometric interpretation,
as shown in Figure 8.9. Note that because the nth roots all have the same modulus (length)
n n
͙r , they will lie on a circle of radius ͙r with center at the origin. Furthermore, the n
roots are equally spaced along the circle, since successive nth roots have arguments that
differ by 2␲͞n.
nth Roots of a Complex Number
We have already found the sixth roots of 1 by factoring and the quadratic formula. Try
solving the same problem using Theorem 8.6 to see if you get the roots shown in Figure
8.10. When Theorem 8.6 is applied to the real number 1, we give the nth roots a special
name—the nth roots of unity.
Figure 8.10
Imaginary
axis
−1 + 3i 1+ 3
i
2 2 2 2
−1 1 Real
axis
−1 − 3i 1− 3
i
2 2 2 2
6th Roots of Unity
EXAMPLE 5 Finding the nth Roots of a Complex Number
Determine the fourth roots of i.

8. 452 CHAPTER 8 COMPLEX VECTOR SPACES
Solution In polar form, we can write i as
␲ ␲
΂
i ϭ 1 cos ϩ i sin
2 2 ΃
so that r ϭ 1, ␪ ϭ ␲͞2. Then, by applying Theorem 8.6, we have
␲͞2 2k␲ ␲͞2 2k␲
i1͞4 ϭ͙1 cos
4
΄ ΂4 ϩ
4 ΃ϩ i sin
4
ϩ ΂
4 ΃΅
␲ k␲ ␲ k␲
ϭ cos ΂ 8 ϩ 2 ΃ ϩ i sin ΂ 8 ϩ 2 ΃.
Setting k ϭ 0, 1, 2, and 3 we obtain the four roots
␲ ␲
z1 ϭ cos ϩ i sin
8 8
5␲ 5␲
z 2 ϭ cos ϩ i sin
8 8
9␲ 9␲
z3 ϭ cos ϩ i sin
8 8
13␲ 13␲
z4 ϭ cos ϩ i sin
8 8
as shown in Figure 8.11.
REMARK: In Figure 8.11 note that when each of the four angles, ␲͞8, 5␲͞8, 9␲͞8,
and 13␲͞8 is multiplied by 4, the result is of the form ͑␲͞2͒ ϩ 2k␲.
Figure 8.11
Imaginary
axis
cos 5π + i sin 5π
8 8
cos π + i sin π
8 8
Real
axis
cos 9π + i sin 9π
8 8
cos 13π + i sin 13π
8 8

9. SECTION 8.3 EXERCISES 453
SECTION 8.3 u EXERCISES
In Exercises 1–4, express the complex number in polar form. 25. 7͑cos 0 ϩ i sin 0͒ 26. 6͑cos␲ ϩ i sin ␲͒
1. Imaginary 2. Imaginary In Exercises 27–34, perform the indicated operation and leave the
axis axis result in polar form.
Real 3i
␲ ␲ ␲ ␲
΄΂ ΃΅΄4΂cos 6 ϩ i sin 6 ΃΅
3
1 2
axis 27. 3 cos ϩ i sin
2 3 3
−1
␲ ␲ ␲ ␲
΄4 ΂cos 2 ϩ i sin 2 ΃΅΄6΂cos 4 ϩ i sin 4 ΃΅
1 3
28.
Real
−2 axis
2 − 2i −1 1
29. ͓0.5͑cos␲ ϩ i sin␲͔͒ ͓ 0.5͑cos͓Ϫ␲͔ ϩ i sin͓Ϫ␲͔͔͒
␲ ␲ 2␲ 2␲
΄΂ ΃΅΄3 ΂cos 3 ΃΅
1
3. Imaginary 4. Imaginary 30. 3 cos ϩ i sin ϩ i sin
axis axis 3 3 3
1 + 3i
3 3 2[cos(2␲͞3) ϩ i sin(2␲͞3)]
2 31.
4[cos(2␲͞9) ϩ i sin(2␲͞9)]
−6 1
Real
2
axis cos(5␲͞3) ϩ i sin(5␲͞3)
−6 −5 −4 −3 −2 1 32.
cos␲ ϩ i sin␲
−2 Real
−3 −1 1 2
axis 12[cos(␲͞3) ϩ i sin(␲͞3)]
33.
3[cos(␲͞6) ϩ i sin(␲͞6)]
In Exercises 5–16, represent the complex number graphically, and
give the polar form of the number. 9[cos(3␲͞4) ϩ i sin(3␲͞4)]
34.
5[cos(Ϫ␲͞4) ϩ i sin(Ϫ␲͞4)]
5. Ϫ2 Ϫ 2i 6. ͙3 ϩ i
7. Ϫ2͑1 ϩ ͙3 i͒ 8. 5
2 ͙͑3 Ϫ i ͒ In Exercises 35–44, use DeMoivre’s Theorem to find the indicated
9. 6i 10. 4 powers of the given complex number. Express the result in standard
form.
11. 7 12. Ϫ2i
13. 1 ϩ 6i 14. 2͙2 Ϫ i 35. ͑1 ϩ i͒ 4 36. ͑2 ϩ 2i͒ 6
38. ͙͑3 ϩ i͒
7
15. Ϫ3 Ϫ i 16. Ϫ4 ϩ 2i 37. ͑Ϫ1 ϩ i͒ 10
␲ ␲
΄΂ ΃΅
3
39. ͑1 Ϫ ͙3i͒
3
In Exercises 17–26, represent the complex number graphically, and 40. 5 cos ϩ i sin
9 9
give the standard form of the number.
5␲ 5␲ 5␲ 5␲
΄΂ ΃΅ ΂ ΃
4 10
␲ ␲ 3␲ 3␲
΂
17. 2 cos
2
ϩ i sin
2 ΃ 18. 5 cos ΂ 4
ϩ i sin
4 ΃ 41. 3 cos
6
ϩ i sin
6
42. cos
4
ϩ i sin
4
␲ ␲ 3␲ 3␲
΄΂ ΃΅ ΄΂ ΃΅
8 4
5␲ 5␲ 7␲ 7␲
΂ ΃ ΂ ΃ ϩ i sin ϩ i sin
3 3 43. 2 cos 44. 5 cos
19. cos ϩ i sin 20. cos ϩ i sin 2 2 2 2
2 3 3 4 4 4
␲ ␲ ␲ ␲ In Exercises 45–56, (a) use DeMoivre’s Theorem to find the indi-
21. 3.75 cos ΂ 4
ϩ i sin
4 ΃ 22. 8 cos ΂ 6
ϩ i sin
6 ΃ cated roots, (b) represent each of the roots graphically, and (c) ex-
press each of the roots in standard form.
3␲ 3␲ 5␲ 5␲ ␲ ␲
΂
23. 4 cos
2
ϩ i sin
2 ΃ 24. 6 cos ΂ 6
ϩ i sin
6 ΃ ΂
45. Square roots: 16 cos ϩ i sin
3 3 ΃

10. 454 CHAPTER 8 COMPLEX VECTOR SPACES
2␲ 2␲ 66. Show that the negative of z ϭ r͑cos␪ ϩ i sin␪͒ is
΂
46. Square roots: 9 cos
3
ϩ i sin
3 ΃ Ϫz ϭ r͓cos͑␪ ϩ ␲͒ ϩ i sin͑␪ ϩ ␲͔͒.
4␲ 4␲ ␲ ␲
47. Fourth roots: 16 cos ΂ 3
ϩ i sin
3 ΃ ΂
67. (a) Let z ϭ r͑cos␪ ϩ i sin␪͒ ϭ 2 cos
6 ΃
ϩ i sin . Sketch z,
6
5␲ 5␲ iz, and z͞i in the complex plane.
΂
48. Fifth roots: 32 cos
6
ϩ i sin
6 ΃ (b) What is the geometric effect of multiplying a complex
49. Square roots: Ϫ25i 50. Fourth roots: 625i number z by i? What is the geometric effect of dividing z
by i?
125 68. (Calculus) Recall that the Maclaurin series for e x, sin x, and
51. Cube roots: Ϫ ͑1 ϩ ͙3i͒
2 cos x are
52. Cube roots: Ϫ4͙2 ͑1 Ϫ i͒ x 2 x3 x 4 . . .
ex ϭ 1 ϩ x ϩ ϩ ϩ ϩ
53. Cube roots: 8 54. Fourth roots: i 2! 3! 4!
55. Fourth roots: 1 56. Cube roots: 1000 x 3 x5 x 7 . . .
sin x ϭ x Ϫ ϩ Ϫ ϩ
3! 5! 7!
In Exercises 57–62, find all the solutions to the given equation and
represent your solutions graphically. x 2 x 4 x6 . . .
cos x ϭ 1 Ϫ ϩ Ϫ ϩ .
2! 4! 6!
57. x4 Ϫ i ϭ 0 58. x 3 ϩ 1 ϭ 0
59. x 5 ϩ 243 ϭ 0 60. x 4 Ϫ 81 ϭ 0 (a) Substitute x ϭ i␪ into the series for e x and show that
61. x 3 ϩ 64i ϭ 0 62. x 4 ϩ i ϭ 0 ei␪ ϭ cos ␪ ϩ i sin ␪.
63. Given two complex numbers z 1 ϭ r1͑cos ␪ 1 ϩ i sin ␪ 1͒ and (b) Show that any complex number z ϭ a ϩ bi can be
z 2 ϭ r2͑cos␪ 2 ϩ i sin ␪ 2 ͒ with z 2 0 prove that expressed in polar form as z ϭ rei␪.
z 1 r1 (c) Prove that if z ϭ rei␪, then z ϭ reϪi␪.
z 2 ϭ r 2 ͓cos͑␪ 1 Ϫ ␪ 2͒ ϩ i sin͑␪ 1 Ϫ ␪ 2͔͒. (d) Prove the amazing formula ei␲ ϭ Ϫ1.
64. Show that the complex conjugate of z ϭ r͑cos␪ ϩ i sin␪͒ is
z ϭ r ͓cos͑Ϫ␪͒ ϩ i sin͑Ϫ␪͔͒.
65. Use the polar form of z and z in Exercise 64 to find the fol-
lowing.
(a) zz (b) z͞z, z 0