Download Section 9.3 – The Complex Plane and De Moivre`s Theorem

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Section 9.3 – The Complex Plane and De Moivre's Theorem
Objective 1:
Understanding the Complex Plane.
In this chapter, we will be working with complex numbers. Recall that a
complex number is in the form z = x + yi, where x is the real part of z and y
is the imaginary part of z and i = −1 . We can use the coordinate axes,
called the complex plane, to represent a complex number graphically. The
x-axis is the real axis and the y-axis is the imaginary axis. Thus, if x = 0 and
y ≠ 0, then the complex number will lie on the y-axis and it will be a strictly
imaginary number while if y = 0, then the complex number will lie on the xaxis and it will be a strictly real number. For all other complex numbers, z
will be represented as a point (x, y) on the complex plane. The distance
from the origin to the point (x, y) representing complex number is equal to
x2 + y 2 . We will call this the magnitude or modulus of z.
Imaginary Axis
y
z = x + yi
Real Axis
x
Definition
Let z = x + yi be a complex number. The magnitude or modulus of z,
denote |z|, is defined as the distance from the origin to the point (x, y) or
|z| =
x2 + y 2
This is also called the absolute value of z.
Recall that if z = x + yi, then the complex conjugate of z = x – yi. If we
multiply a complex number with its conjugate, we get:
(x + yi)(x – yi) = x2 – y2i2 = x2 – y2(– 1) = x2 + y2
Theorem
Let z = x + yi and let z be the complex conjugate of z. Then
|z| = z • z
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Objective 2:
Convert a Complex Number from Rectangular to Polar
Form.
Recall that if a point (r, θ) was in polar form, we were able to convert it into
rectangular form by using x = rcos(θ) and y = rsin(θ). Thus, for a complex
number in rectangular, z = x + yi, we can replace x by rcos(θ) and y by
rsin(θ) to write the complex number in polar form. To make a representation
of a complex number in polar form unique, we will restrict r to be greater
than or equal to zero and θ to be in the interval [0, 2π).
Definition
If r ≥ 0 and 0 ≤ θ < 2π, then the complex number z = x + yi can be written in
polar form as:
z = [rcos(θ)] + [rsin(θ)]i = r[cos(θ) + isin(θ)]
where r = |z| and the θ is called the argument of z.
Plot the following numbers in the complex plane and then write the
number in polar form:
Ex. 1a
5 – 5i
Ex. 1b
– 2 – ( 3 )i
Ex. 1c
– 3 +i
Ex. 1d
2 + 2i
Solution:
a)
First, we plot the point (5, – 5) on
the complex plane. This point is
in quadrant IV. Next, we will
calculate r = |z|:
r2 = (5)2 + (– 5)2 = 50
r = 50 = 5 2
–5
Since sin(θ) =
y
r
=
−5
5 2
=–
2
2
5
5
,
then our reference angle is
θR = 45˚. Because the number
is in quadrant IV, then θ = 360˚ – 45˚ = 315˚.
Plugging r and θ into the polar form, we get:
z = 5 2 [cos(315˚) + isin(315˚)]
–5
z = 5 – 5i
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b)
First, we plot the point (– 2, – 3 ) on
the complex plane. This point is
in quadrant III. Next, we will
calculate r = |z|:
r2 = (– 2)2 + (– 3 )2 = 7
r=
7
Since sin(θ) =
y
r
then θ = sin – 1(–
=
− 3
7
21
7
=–
21
7
5
–5
,
5
– 2 – ( 3 )i
) = – 40.893…˚
–5
Thus, our reference angle is
θR ≈ 40.89˚. Because the number
is in quadrant III, then θ = 180˚ + 40.89˚ = 220.89˚.
Plugging r and θ into the polar form, we get:
z = 7 [cos(220.89˚) + isin(220.89˚)]
c)
First, we plot the point (– 3 , 1) on
the complex plane. This point is
in quadrant II. Next, we will
–
calculate r = |z|:
r2 = (– 3 )2 + (1)2 = 4
r=
3 +i
–5
4 =2
Since sin(θ) =
5
y
r
=
1
,
2
then our reference angle is
θR = 30˚. Because the number
is in quadrant II, then θ = 180˚ – 30˚ = 150˚.
Plugging r and θ into the polar form, we get:
z = 2[cos(150˚) + isin(150˚)]
d)
5
First, we plot the point (2, 2) on
the complex plane. This point is
in quadrant I. Next, we will
calculate r = |z|:
r2 = (2)2 + (2)2 = 8
r= 8 =2 2
–5
5
(2, 2)
–5
5
–5
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Since sin(θ) =
y
r
=
2
2 2
=
2
2
,
then our reference angle is
θR = 45˚. Because the number
is in quadrant I, then θ = 45˚.
Plugging r and θ into the polar form, we get:
z = 2 2 [cos(45˚) + isin(45˚)]
Write the following numbers in rectangular form:
5π
5π
Ex. 2a
2[cos(
) + isin(
)]
6
6
Ex. 2b
0.4[cos(200˚) + isin(200˚)]
Solution:
a)
We simply evaluate the trigonometric functions and then
distribute:
2[cos(
b)
5π
6
) + isin(
5π
6
)] = 2[–
3
2
+
1
i]
2
=–
3 +i
0.4[cos(200˚) + isin(200˚)] = 0.4[– 0.93969… – 0.34202…i]
≈ – 0.3759 – 0.1368i
Objective 3:
The Product and Quotient of Two Complex Numbers in
Polar Form.
We will now examine how to multiply or divide two complex numbers in
polar form.
Product & Quotient Theorem for Complex Numbers in Polar Form
Let z1 = r1[cos(θ1) + isin(θ1)] and z2 = r2[cos(θ2) + isin(θ2)].
1)
Then z1•z2 = r1•r2[cos(θ1 + θ2) + isin(θ1 + θ2)] and
2)
If z2 ≠ 0, then
z1
z2
=
r1
r2
[cos(θ1 – θ2) + isin(θ1 – θ2)]
Proof:
1) z1•z2 = r1[cos(θ1) + isin(θ1)]• r2[cos(θ2) + isin(θ2)] (group r1 & r2 together)
= r1•r2[cos(θ1) + isin(θ1)][cos(θ2) + isin(θ2)]
(FOIL)
= r1•r2[cos(θ1)cos(θ2) + icos(θ1)sin(θ2) + isin(θ1)cos(θ2) + i2sin(θ1)sin(θ2)]
= r1•r2[cos(θ1)cos(θ2) + icos(θ1)sin(θ2) + isin(θ1)cos(θ2) – sin(θ1)sin(θ2)]
(group the real parts together and the imaginary parts together)
= r1•r2[cos(θ1)cos(θ2) – sin(θ1)sin(θ2) + icos(θ1)sin(θ2) + isin(θ1)cos(θ2)]
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(but cos(θ1 + θ2) = cos(θ1)cos(θ2) – sin(θ1)sin(θ2) and
sin(θ1 + θ2) = cos(θ1)sin(θ2) + sin(θ1)cos(θ2))
= r1•r2[cos(θ1 + θ2) + isin(θ1 + θ2)]
2)
=
=
=
z1
r [cos(θ1) +isin(θ1 )]
= 1
(multiply
z2
r2 [cos(θ2 ) +i sin(θ2 )]
r1[cos(θ1) +isin(θ1 )] [cos(θ2 ) −isin(θ2 )]
•
r2 [cos(θ2 ) +i sin(θ2 )] [cos(θ2 ) −isin(θ2 )]
the top & bottom by cos(θ2) – isin(θ2))
(expand)
r1[cos(θ1)cos(θ2 ) − icos(θ1)sin(θ2 ) + isin(θ1)cos(θ2 ) − i2sin(θ1)sin(θ2 )]
r2 [cos2 (θ2 )− i2 sin2 (θ2 )]
r1[cos(θ1)cos(θ2 ) − icos(θ1)sin(θ2 ) + isin(θ1)cos(θ2 ) +sin(θ1 )sin(θ2 )]
r2 [cos2 (θ 2 )+ sin2 (θ2 )]
(but cos2(θ2) + sin2(θ2) = 1)
=
=
r1[cos(θ1)cos(θ2 ) − icos(θ1)sin(θ2 ) + isin(θ1)cos(θ2 ) +sin(θ1 )sin(θ2 )]
r2 [1]
r1
[cos(θ1)cos(θ2) – icos(θ1)sin(θ2) + isin(θ1)cos(θ2) + sin(θ1)sin(θ2)]
r2
(group the real parts together and the imaginary parts together)
=
r1
r2
[cos(θ1)cos(θ2) + sin(θ1)sin(θ2) – icos(θ1)sin(θ2) + isin(θ1)cos(θ2)]
(but cos(θ1 – θ2) = cos(θ1)cos(θ2) + sin(θ1)sin(θ2) and
sin(θ1 – θ2) = cos(θ1)sin(θ2) – sin(θ1)cos(θ2))
=
r1
r2
[cos(θ1 – θ2) + isin(θ1 – θ2)].
Thus, the proof is complete.
Given z = 2[cos(80˚) + isin(80˚)] and w = 6[cos(200˚) + isin(200˚)], find:
z
Ex. 3a
zw
Ex. 3b
w
Solution:
a)
zw = r1•r2[cos(θ1 + θ2) + isin(θ1 + θ2)]
= (2)(6)[cos(80˚ + 200˚) + isin(80˚ + 200˚)]
= 12[cos(280˚) + isin(280˚)]
b)
z
w
=
=
2
[cos(80˚ – 200˚) + isin(80˚ –
6
1
[cos(– 120˚) + isin(– 120˚)]
3
=
r1
r2
[cos(θ1 – θ2) + isin(θ1 – θ2)]
200˚)]
But, we need an angle in the interval [0, 360˚), so add 360˚:
– 120˚ + 360˚ = 240˚
=
1
[cos(240˚)
3
+ isin(240˚)]
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Objective 4:
Use De Moivre's Theorem
Raising a complex number to a power and expanding in rectangular
coordinates can be a very tedious task. For instance, if we were asked to
evaluate (1 – 5 i)12, we would have to multiply 12 factors of (1 – 5 i).
We need to find a better way of simplifying this problem. The key will be to
first convert the complex number into polar form and then use a theorem
called De Moivre's Theorem to evaluate the expression.
€
De Moivre's Theorem
Let z = r[cos(θ) + isin(θ)] and n be a natural integer. Then
zn = rn[cos(nθ) + isin(nθ)]
Proof:
To prove this, we will need to use a method called mathematical induction.
In mathematical induction, you first show that the statement is true for
some initial value of n, usually n = 1. You then assume the statement is
true for value k and then show it to be true for k + 1.
Part I: n = 1
z(1) = r(1)[cos((1)θ) + isin((1)θ)] = r[cos(θ) + isin(θ)] true
Part II: Assume it is true for n = k, show it is true for n = k + 1.
For n = k, rk = rk[cos(kθ) + isin(kθ)] is assumed to be true.
zk + 1 = zk•z = rk[cos(kθ) + isin(kθ)]•r[cos(θ) + isin(θ)]
(use the product and quotient theorem)
= rk•r[cos(kθ + θ) + isin(kθ + θ)]
= rk + 1[cos([k + 1]θ) + isin([k + 1]θ)]
Now, we get the domino effect:
Part I says the statement is true for n = 1
Part II implies that if it is true for n = 1, then it is true for n = 2.
But, part II also implies that if it is true for n = 2, then it is true for n = 3.
But, this implies that if it is true for n = 3, then it is true for n = 4, and so
on.
Hence, by mathematical induction, this statement is true for any natural
number n.
Simplify. Write your answer in both polar and rectangular form:
Ex. 4a
[3(cos(280˚) + isin(280˚)]4
Ex. 4b
(1 – 5 i)12
Solution:
a)
[3(cos(280˚) + isin(280˚)]4
(use De Moivre's Theorem)
4
= 3 (cos(4•280˚) + isin(4•280˚))
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= 81(cos(1120˚) + isin(1120˚))
But the argument needs to be in [0, 360˚).
1120˚ – 3(360˚) = 40˚
= 81(cos(40˚) + isin(40˚))
Polar Form
= 81(0.76604… + (0.64278…)i)
≈ 62.05 + 52.07i
Rectangular Form
b)
1–
(convert 1 –
5 i)12
5 i is in quadrant IV
r=
(1)2 +(− 5 )2 =
(1 –
sin(θ) =
y
r
=
− 5
6
5 i to polar form)
6
, so θ = sin – 1(
− 5
6
) ≈ – 65.91˚
Thus, θR ≈ 65.91˚ and so θ = 360 – 65.91 = 294.09˚.
Plugging in, we get:
(1 – 5 i)12 = ( 6 [cos(294.09˚) + isin(294.09˚)])12
(apply De Moirve's Theorem)
= ( 6 )12[cos(12•294.09˚) + isin(12•294.09˚)]
= 46656[cos(3529.14˚) + isin(3529.14˚)]
= 46656[cos(289.14˚) + isin(289.14˚)] Polar Form
= 46656[0.32784… – 0.94473…i]
= 15296 – 44077.371…i
= 15296 – 19712 5 i
Objective 5:
Find Complex Roots
Now, we will consider finding the complex nth root of a number. For a
number z be a complex nth root of a complex number w, the following
equation has to be true:
zn = w
Finding Complex Roots Theorem
Let w = r(cos(θ) + isin(θ)) be a nonzero complex number and let n ≥ 2 be a
natural number. Then the equation zn = w has n distinct complex roots
given by:
θ
2kπ
θ
2kπ
n
zk = r [cos( +
) + isin( +
)] where k = 0, 1, 2,…, n – 1
n
n
n
Find all the complex fourth roots of:
Ex. 5
– 27 – 27i
n
208
Solution:
First, we will rewrite the number in polar form. The number is in
quadrant III:
r2 = (– 27)2 + (– 27)2 = 1458
r = 1458 = 27 2
sin(θ) =
y
r
−27
=
27 2
=–
2
2
, so our reference angle is θR = 45˚.
Thus, θ = 180˚ + 45˚ = 225˚.
Hence, the number in polar form is:
27 2 [cos(225˚) + isin(225˚)]
Now, apply the formula using n = 4:
But,
zk =
4
zk =
4
4
27 2 [cos(
225o
4
+
2k(180o )
4
) + isin(
225o
4
+
2k(180o )
4
1458 [cos(56.25˚ + 90˚k) + isin(56.25˚ + 90˚k)]
1458 = ((1458)1/2)1/4 = (1458)1/8 =
8
1458
8
zk = 1458 [cos(56.25˚ + 90˚k) + isin(56.25˚ + 90˚k)]
for k = 0, 1, 2, and 3.
Thus, the four roots are:
z0 =
8
1458 [cos(56.25˚ + 90˚(0)) + isin(56.25˚ + 90˚(0))]
=
z1 =
8
8
8
1458 [cos(146.25˚) + isin(146.25˚)]
8
1458 [cos(236.25˚) + isin(236.25˚)], and
1458 [cos(56.25˚ + 90˚(3)) + isin(56.25˚ + 90˚(3))]
=
€
8
1458 [cos(56.25˚ + 90˚(2)) + isin(56.25˚ + 90˚(2))]
=
z3 =
1458 [cos(56.25˚) + isin(56.25˚)],
1458 [cos(56.25˚ + 90˚(1)) + isin(56.25˚ + 90˚(1))]
=
z2 =
8
8
1458 [cos(326.25˚) + isin(326.25˚)].
)]