Download Complex Numbers in Polar Form

Complex Numbers
in Polar Form;
DeMoivre’s Theorem
The Complex Plane
We know that a real number can be represented as a point on a number line.
By contrast, a complex number z = a + bi is represented as a point (a, b) in a
coordinate plane, shown below. The horizontal axis of the coordinate plane is
called the real axis. The vertical axis is called the imaginary axis. The
coordinate system is called the complex plane. Every complex number
corresponds to a point in the complex plane and every point in the complex
plane corresponds to a complex number.
Imaginary
axis
b
z = a + bi
Real axis
a
Text Example
Plot in the complex plane:
a. z = 3 + 4i
b. z = -1 – 2i
Solution
c. z = -3
•
We plot the complex number z = 3 + 4i
the same way we plot (3, 4) in the
rectangular coordinate system. We
move three units to the right on the real
axis and four units up parallel to the
imaginary axis.
•
The complex number z = -1 – 2i
corresponds to the point (-1, -2) in the
rectangular coordinate system. Plot the
complex number by moving one unit
to the left on the real axis and two units
down parallel to the imaginary axis.
d. z = -4i
5
4
3
2
1
-5 -4 -3 -2
z = -1 – 2i
z = 3 + 4i
1 2 3 4 5
-3
-4
-5
1
Text Example cont.
Plot in the complex plane:
a. z = 3 + 4i
b. z = -1 – 2i
Solution
•
•
c. z = -3
Because z = -3 = -3 + 0i, this complex
number corresponds to the point (-3, 0).
We plot –3 by moving three units to the
left on the real axis.
Because z = -4i = 0 – 4i, this complex
number corresponds to the point (0, -4).
We plot the complex number by moving
three units down on the imaginary axis.
d. z = -4i
z = -3
-5 -4 -3 -2
z = -1 – 2i
z = 3 + 4i
5
4
3
2
1
1 2 3 4 5
-3
-4
-5
z = -4i
The Absolute Value of a
Complex Number
• The absolute value of the complex number
a + bi is
2
2
z = a + bi = a + b
Example
• Determine the absolute value of z=2-4i
Solution:
z = a + bi = a 2 + b 2
= 2 2 + (−4) 2 = 4 + 16
= 20 = 2 5
2
Polar Form of a Complex
Number
The complex number a + bi is written in polar form as
z = r (cos θ + i sin θ )
2
2
where a = r cos θ , b = r sin θ , r = a + b
and
tan θ=b/a The value of r is called the modulus
(plural: moduli) of the complex number z, and the
angle θ is called the argument of the complex
number z, with 0 < θ < 2π
Text Example
Plot z = -2 – 2i in the complex plane. Then write z in polar form.
Solution The complex number z = -2 – 2i, graphed below, is in rectangular
form a + bi, with a = -2 and b = -2. By definition, the polar form of z is r(cos
θ + i sin θ ). We need to determine the value for r and the value for θ ,
included in the figure below.
Imaginary
axis
2
è
2
-2
Real
axis
r
-2
z = -2 – 2i
Text Example cont.
Solution
r = a + b = (−2) + (−2) = 4 + 4 = 8 = 2 2
2
2
2
tan θ =
2
b −2
=
=1
a −2
Since tan π/4 = 1, we know that θ lies in quadrant III. Thus,
θ =π +
π
4
+
4π π 5π
+ =
4
4
4
z = r(cosθ + i sinθ ) = 2 2(cos
5π
5π
+ i sin )
4
4
3
Product of Two Complex
Numbers in Polar Form
Let z1 = r1 (cos θ1+ i sin θ 1) and
z2 = r2 (cos θ 2 + i sin θ 2) be two
complex numbers in polar form. Their
product, z1z2, is
z1z2 = r1 r2 (cos (θ 1 + θ 2) + i sin (θ 1 + θ 2))
To multiply two complex numbers,
multiply moduli and add arguments.
Text Example
Find the product of the complex numbers. Leave the answer in polar form.
z1 = 4(cos 50º + i sin 50º) z2 = 7(cos 100º + i sin 100º)
Solution
z1z2
Form the product of the given
= [4(cos 50º + i sin 50º)][7(cos 100º + i sin 100º)]
numbers.
= (4 · 7)[cos (50º + 100º) + i sin (50º + 100º)]
Multiply moduli and add
arguments.
= 28(cos 150º + i sin 150º)
Simplify.
Quotient of Two Complex
Numbers in Polar Form
Let z1 = r1 (cos θ1 + i sin θ1) and z2 = r2 (cos
θ2 + i sin θ2) be two complex
numbers in polar form. Their quotient, z1/z2, is
z1 r1
= [cos(θ1 − θ 2 ) + i sin(θ1 − θ 2 )]
z 2 r2
To divide two complex numbers, divide
moduli and subtract arguments.
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DeMoivre’s Theorem
• Let z = r (cos θ + i sin θ) be a complex
numbers in polar form. If n is a positive
integer, z to the nth power, zn, is
z n = [r (cos θ + i sin θ )]n
= r n (cos nθ + i sin nθ )
Text Example
Find [2 (cos 10º + i sin 10º)]6. Write the answer in rectangular form a + bi.
Solution By DeMoivre’s Theorem,
[2 (cos 10º + i sin 10º)]6
= 26[cos (6 · 10º) + i sin (6 · 10º)]
Raise the modulus to the 6th power and multiply
the argument by 6.
= 64(cos 60º + i sin 60º)
Simplify.
1
3
= 64  + i

2 
2
Write the answer in rectangular form.
= 32 + 32 3i
Multiply and express the answer in a + bi form.
DeMoivre’s Theorem for Finding
Complex Roots
• Let ω=r(cosθ+isinθ) be a complex number
in polar form. If ω≠0, ω has n distinct
complex nth roots given by the formula
  θ + 360k 
 θ + 360k 
z k = n r cos
 + i sin 

n
n



 
where k = 0,1,2,3,..., n − 1
5
Example
• Find all the complex fourth roots of
81(cos60º+isin60º)
Solution:
  θ + 360 k 
 θ + 360 k 
z k = n r cos
 + i sin 

n
n



 
  60 + 360 * 0 
 60 + 360 * 0 
= 4 81 cos

 + i sin 
4
4



 
= 3(cos15o + i sin 15o )
Example cont.
• Find all the complex fourth roots of
81(cos60º+isin60º)
Solution:
  60 + 360 *1 
 60 + 360 *1 
= 4 81 cos
 + i sin 

4
4



 
o
o
= 3(cos 105 + i sin 105 )
= 3(cos 195o + i sin 195o )
= 3(cos 285o + i sin 285o )
Complex Numbers
in Polar Form;
DeMoivre’s
Theorem
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