Download Module 10 Lesson 5: Polar Form of a Complex Number

Module 10 Lesson 5: Polar Form of a Complex Number
In the previous Lesson, you learned what a complex number is, how to write complex
numbers in standard, how to add, subtract, multiply, and divide complex numbers, and
even how to plot them in the complex plane.
Next, we will look at how to express complex number in polar
form. The polar form of a complex number uses the modulus
and an angle to describe a complex number’s location in the
plane.
Before we begin rewriting complex numbers in polar form, let’s
first look at how to find the absolute value of a complex
number. This skill will help us later on when determining a
complex number’s polar form.
Note: Complex form is also
known as rectangular form
and polar form is also known
as trigonometric form if you
happen to search for other
resources to support you in
this lesson.
The Absolute Value (or Modulus) of a Complex Number
Say we have a complex number, which we will call z.
𝑧 = 𝑎 + 𝑏𝑖
Then the absolute value (or modulus) of z is defined as
|𝑧| = |𝑎 + 𝑏𝑖| = √𝑎2 + 𝑏 2
The absolute value of a complex number is the distance between the origin (0, 0) and
the points (a, b). We can see how this is true by the application of the Pythagorean
Theorem. See the image below.
1
Ex. Find the absolute value of 4 − 7𝑖.
For this complex number, 𝑎 = 4 𝑎𝑛𝑑 𝑏 = −7. So we have…
|4 − 7𝑖| = √42 + (−7)2 = √16 + 49 = √65
Now, if the complex number 𝑎 + 𝑏𝑖 is a real number (that is, if b = 0), then the
definition given above remains true.
Ex. Find the absolute value of −6.
Here, 𝑎 = −6 𝑎𝑛𝑑 𝑏 = 0.
|−6| = √(−6)2 + 02 = √36 = 6
Polar Form of a Complex Number
We already know how to add, subtract, multiply and divide complex numbers. In order
to work with powers and roots of complex numbers, it is best to work with the polar (or
trigonometric) form of a complex number.
Recall that standard form of a complex number is 𝑎 + 𝑏𝑖. As mentioned before, the
polar form of a complex number relates the location of the number in terms of an
angle. Here, we will call this angle θ. See the figure below.
2
Using your knowledge of right triangle Trigonometry (from both your previous
Geometry course as well as Module 4), note that the following two statements are true
for this triangle:
𝑎
𝑏
cos 𝜃 = 𝑟 and sin 𝜃 = 𝑟
We will rewrite the “a” and “b” values found in our complex form as:
𝑎 = 𝑟𝑐𝑜𝑠𝜃 and 𝑏 = 𝑟𝑠𝑖𝑛𝜃
𝑟 = √𝑎2 + 𝑏 2
𝑏
and we can find 𝜃, by tan 𝜃 = 𝑎
Therefore, the Polar Form of a Complex Number is:
𝑎 + 𝑏𝑖 = (𝑟𝑐𝑜𝑠𝜃) + (𝑟𝑠𝑖𝑛𝜃)𝑖
Lastly, we can simplify this expression by factoring out the “r” and letting “𝑖” be written
in front of 𝑠𝑖𝑛𝜃 (so that we make sure we are not taking the sine of θ multiplied by 𝑖).
𝑧 = 𝑟(cos 𝜃 + 𝑖 sin 𝜃)
This is the polar form of a complex number. The “r” is called the modulus and the “θ” is
called the argument. We will refer to these variables as the modulus and argument of
the complex number in this lesson.
Note: Almost all the time, the angle θ will be restricted to the interval 0 ≤ 𝜃 <
2𝜋 since there are actually infinitely many choices for θ. Sometimes it is useful
to use θ < 0.
Now, let’s look at several examples of how to write a complex number in polar form.
3
Ex. Write the complex number 𝑧 = 1 − √3𝑖 in polar form with 0 ≤ 𝜃 < 2𝜋.
Here, 𝑎 = 1 𝑎𝑛𝑑 𝑏 = −√3. So 𝑟 = √12 + (−√3)2 = √1 + 3 = √4 = 2.
𝑏
In order to find 𝜃, we will look at the value for tan 𝜃 = 𝑎.
tan 𝜃 =
−√3
= −√3
1
Recall your knowledge from The Unit Circle (Module 4) and Inverse Trig
Functions (Module 5). It is helpful to draw a quick sketch of this complex
number and look at the Quadrant in which the complex number lies. Here, 𝑧 =
1 − √3𝑖 lies in Quadrant IV.
Note: This image is
not necessarily drawn
to scale. It’s just a
sketch to show the
location of z.
𝜋
𝜋
Solving for 𝜃, we find a reference angle by 𝜃 ′ = tan−1 −√3 = − 3 because at – 3 ,
𝜋
tan − 3 =
4
−√3/2
1/2
= −√3.
𝜋
However, we must write 𝜃 such that 0 ≤ 𝜃 < 2𝜋. Therefore, 𝜃 ′ = − 3 also
corresponds to 𝜃 =
5𝜋
3
. See the figure below.
Using these values for r and 𝜃, let’s rewrite our complex number in polar form.
𝑧 = 𝑟(cos 𝜃 + 𝑖 sin 𝜃)
5𝜋
5𝜋
= 2 (cos
+ 𝑖 sin )
3
3
So the polar form for 1 − √3𝑖 is 2 (cos
5
5𝜋
3
+ 𝑖 sin
5𝜋
3
).
Ex. Find the polar form with 0 ≤ 𝜃 < 2𝜋 for the complex number −3 − 4𝑖.
Here, 𝑎 = −3 𝑎𝑛𝑑 𝑏 = −4 and the modulus is 𝑟 = √(−3)2 + (−4)2 =
√9 + 16 = √25 = 5.
−4
4
Next, find 𝜃 by tan 𝜃 = −3 = 3. This isn’t a “simple” value that lets us use our
Unit Circle knowledge. Instead, let’s use our calculator to find the value of 𝜃.
4
𝜃 = tan−1 3 ≈ 0.927
The calculation above gave us an angle that lies in Quadrant I.
*Remember to have
your calculator in
Radian Mode to find the
value of 𝜃.
Because tangent is also positive in Quadrant III, and this is in fact the quadrant in
which our complex number lies, we must conclude that 𝜃 = 𝜋 + 0.927 ≈ 4.069.
Now we can write our complex number in polar form.
−3 − 4𝑖 ≈ 5(cos 4.069 + 𝑖 sin 4.069)
6
Polar From to Complex Form
In the previous example, we transformed a number in complex form to a number in
polar form. Here, let’s take a complex number in polar form and transform it to
standard form 𝑎 + 𝑏𝑖. This is often much easier than changing a complex number in
standard form to polar form. 
Ex. Write the following polar form of a complex number in standard form:
𝜋
𝜋
√8 [cos (− ) + 𝑖 sin (− )]
3
3
𝜋
𝜋
First, find cos (− 3 ) and sin (− 3 ).
𝜋
1
𝜋
cos (− 3 ) = 2 and sin (− 3 ) = −
√3
2
We can also simplify √8 = 2√2.
Substitute these values into the polar form of the complex number.
𝜋
𝜋
1 √3
𝑖)
√8 [cos (− ) + 𝑖 sin (− )] = 2√2 ( −
3
3
2
2
And simplify.
1
2√2 (2 −
7
√3
𝑖)
2
= √2 − √6𝑖
Standard form of our
complex number!
Multiplying Complex Numbers in Polar Form
In Lesson 4 we learned how to multiply and divide complex numbers in standard form.
Here in Lesson 5, we will learn how to multiply and divide complex numbers in polar
form.
There are two formulas involved. Let’s call two complex numbers 𝑧1 and 𝑧2 with
𝑧1 = 𝑟1 (cos 𝜃1 + 𝑖 sin 𝜃1 ) and 𝑧2 = 𝑟2 (cos 𝜃2 + 𝑖 sin 𝜃2 )
The Product and Quotient formulas are as follows:
𝑧1 ∙ 𝑧2 = 𝑟1 ∙ 𝑟2 [cos(𝜃1 + 𝜃2 ) + 𝑖 sin(𝜃1 + 𝜃2 )]
𝑧1 𝑟1
= [cos(𝜃1 − 𝜃2 ) + 𝑖 sin(𝜃1 − 𝜃2 )]
𝑧2 𝑟2
Ex. Express the product of 𝑧1 and 𝑧2 in standard form given
𝜋
𝜋
𝜋
𝜋
4
4
3
3
𝑧1 = 25√2 (cos − + 𝑖 sin − ) and 𝑧2 = 14 (cos + 𝑖 sin )
Let’s apply the formula: 𝑧1 ∙ 𝑧2 = 𝑟1 ∙ 𝑟2 [cos(𝜃1 + 𝜃2 ) + 𝑖 sin(𝜃1 + 𝜃2 )].
𝜋
So we have 𝑧1 ∙ 𝑧2 = 25√2 ∙ 14 [cos (− 4 +
𝜋
3
𝜋
𝜋
) + 𝑖 sin (− 4 + 3 )]
= 350√2 [cos (
𝜋
𝜋
) + 𝑖 sin ( )]
12
12
And using our calculator to find these values:
= 478.11 + 128.11𝑖
8
Note: When adding radian
measures, don’t forget to find
common denominators:
𝜋 𝜋
𝜋 3 𝜋 4
− + =− ∙ + ∙
4 3
4 3 3 4
3𝜋 4𝜋
=−
+
12 12
𝜋
=
12
Ex. Express the quotient of 𝑧1 and 𝑧2 in standard form. Use the same values as those in
the previous example.
𝑧
𝑟
Let’s apply the formula 𝑧1 = 𝑟1 [cos(𝜃1 − 𝜃2 ) + 𝑖 sin(𝜃1 − 𝜃2 )].
2
2
𝑧1 25√2
𝜋 𝜋
𝜋 𝜋
=
[cos (− − ) + 𝑖 sin (− − )]
𝑧2
14
4 3
4 3
=
25√2
7𝜋
7𝜋
[cos (− ) + 𝑖 sin (− )]
14
12
12
And using our calculator (in radian mode) to find these values:
= −0.654 − 2.440𝑖
Reciprocals of Complex Numbers in Polar Form
Lastly, we will learn how to evaluate the reciprocal of a complex number in polar form.
1
Recall that the reciprocal of 𝑥 is 𝑥. Similarly, we can find reciprocals of complex
numbers.
Given any complex number in standard form, we the reciprocal of the complex number
in polar form will be:
1 1
= [cos(−𝜃) + 𝑖 sin(−𝜃)]
𝑧 𝑟
𝜋
𝜋
Ex. Find the reciprocal of 𝑧 = 4 (cos 3 + 𝑖 sin 3 )
Using our formula for the reciprocal,
1 1
= [cos(−𝜃) + 𝑖 sin(−𝜃)]
𝑧 𝑟
1
𝜋
𝜋
= [cos (− ) + 𝑖 sin (− )]
4
3
3
9
𝜋
If asked to write this number in standard form, then we can evaluate cos (− 3 ) and
𝜋
sin (− 3 ) to end up with
=
1
𝜋
𝜋
[cos (− ) + 𝑖 sin (− )]
4
3
3
1 1 √3
= [ −
𝑖]
4 2
2
=
10
1 √3
−
𝑖
8
8