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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 Note: Complex form is also form. The polar form of a complex number uses the modulus known as rectangular form and an angle to describe a complex number’s location in the and polar form is also known plane. as trigonometric form if you happen to search for other Before we begin rewriting complex numbers in polar form, let’s resources to support you in first look at how to find the absolute value of a complex this lesson. number. This skill will help us later on when determining a complex number’s polar form. The Absolute Value (or Modulus) of a Complex Number Say we have a complex number, which we will call z. 𝑧 = π‘Ž + 𝑏𝑖 Then the absolute value (or modulus) of z is defined as 𝑧 = π‘Ž + 𝑏𝑖 =
π‘Ž! + 𝑏 ! 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 βˆ’ 7𝑖. For this complex number, π‘Ž = 4 π‘Žπ‘›π‘‘ 𝑏 = βˆ’7. So we have… 4 βˆ’ 7𝑖 = 4! + (βˆ’7)! = 16 + 49 = 65 Now, if the complex number π‘Ž + 𝑏𝑖 is a real number (that is, if b = 0), then the definition given above remains true. Ex. Find the absolute value of βˆ’6. Here, π‘Ž = βˆ’6 π‘Žπ‘›π‘‘ 𝑏 = 0. βˆ’6 = (βˆ’6)! + 0! = 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 π‘Ž + 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 ΞΈ. 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: !
!
cos πœƒ = ! and sin πœƒ = ! We will rewrite the β€œa” and β€œb” values found in our complex form as: π‘Ž = π‘Ÿπ‘π‘œπ‘ πœƒ and 𝑏 = π‘Ÿπ‘ π‘–π‘›πœƒ π‘Ÿ = π‘Ž! + 𝑏 ! !
and we can find πœƒ, by tan πœƒ = ! Therefore, the Polar Form of a Complex Number is: π‘Ž + 𝑏𝑖 = π‘Ÿπ‘π‘œπ‘ πœƒ + π‘Ÿπ‘ π‘–π‘›πœƒ 𝑖 Lastly, we can simplify this expression by factoring out the β€œr” and letting β€œπ‘–β€ be written in front of π‘ π‘–π‘›πœƒ (so that we make sure we are not taking the sine of ΞΈ multiplied by 𝑖 ). 𝑧 = π‘Ÿ(cos πœƒ + 𝑖 sin πœƒ) This is the polar form of a complex number. The β€œr” is called the modulus and the β€œΞΈβ€ 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 ΞΈ will be restricted to the interval 0 ≀ πœƒ <
2πœ‹ since there are actually infinitely many choices for ΞΈ. Sometimes it is useful to use ΞΈ < 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 βˆ’ 3𝑖 in polar form with 0 ≀ πœƒ < 2πœ‹. Here, π‘Ž = 1 π‘Žπ‘›π‘‘ 𝑏 = βˆ’ 3. So π‘Ÿ =
1! + (βˆ’ 3)! = 1 + 3 = 4 = 2. !
In order to find πœƒ, we will look at the value for tan πœƒ = !. βˆ’ 3
tan πœƒ =
= βˆ’ 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 βˆ’ 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. !
!!
Solving for πœƒ, we find a reference angle by πœƒ = tan
!
tan βˆ’ ! =
4 ! !/!
!/!
= βˆ’ 3. !
!
βˆ’ 3 = βˆ’ ! because at – ! , !
However, we must write πœƒ such that 0 ≀ πœƒ < 2πœ‹. Therefore, πœƒ ! = βˆ’ ! also corresponds to πœƒ =
!!
!
. See the figure below. Using these values for r and πœƒ, let’s rewrite our complex number in polar form. 𝑧 = π‘Ÿ cos πœƒ + 𝑖 sin πœƒ 5πœ‹
5πœ‹
= 2 cos
+ 𝑖 sin
3
3
So the polar form for 1 βˆ’ 3𝑖 is 2 cos
5 !!
!
+ 𝑖 sin
!!
!
. Ex. Find the polar form with 0 ≀ πœƒ < 2πœ‹ for the complex number βˆ’3 βˆ’ 4𝑖. Here, π‘Ž = βˆ’3 π‘Žπ‘›π‘‘ 𝑏 = βˆ’4 and the modulus is π‘Ÿ =
(βˆ’3)! + (βˆ’4)! =
9 + 16 = 25 = 5. !!
!
Next, find πœƒ by tan πœƒ = !! = !. This isn’t a β€œsimple” value that lets us use our Unit Circle knowledge. Instead, let’s use our calculator to find the value of πœƒ. *Remember to have !
πœƒ = tan!! ! β‰ˆ 0.927 your calculator in Radian Mode to find the The calculation above gave us an angle that lies in Quadrant I. value of πœƒ. 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 πœƒ = πœ‹ + 0.927 β‰ˆ 4.069. Now we can write our complex number in polar form. βˆ’3 βˆ’ 4𝑖 β‰ˆ 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 π‘Ž + 𝑏𝑖. This is often much easier than changing a complex number in standard form to polar form. J Ex. Write the following polar form of a complex number in standard form: πœ‹
πœ‹
8 cos βˆ’ + 𝑖 sin βˆ’
3
3
!
!
First, find cos βˆ’ ! and sin βˆ’ ! . !
!
!
cos βˆ’ ! = ! and sin βˆ’ ! = βˆ’
!
!
We can also simplify 8 = 2 2. Substitute these values into the polar form of the complex number. 8 cos βˆ’
πœ‹
πœ‹
+ 𝑖 sin βˆ’
3
3
=2 2
1
3
βˆ’
𝑖 2
2
And simplify. 2 2
7 !
βˆ’
!
!
!
𝑖 = 2 βˆ’ 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 𝑧! and 𝑧! with 𝑧! = π‘Ÿ! (cos πœƒ! + 𝑖 sin πœƒ! ) and 𝑧! = π‘Ÿ! (cos πœƒ! + 𝑖 sin πœƒ! ) The Product and Quotient formulas are as follows: 𝑧! βˆ™ 𝑧! = π‘Ÿ! βˆ™ π‘Ÿ! [cos πœƒ! + πœƒ! + 𝑖 sin πœƒ! + πœƒ! ] 𝑧! π‘Ÿ!
= [cos πœƒ! βˆ’ πœƒ! + 𝑖 sin πœƒ! βˆ’ πœƒ! ] 𝑧! π‘Ÿ!
Ex. Express the product of 𝑧! and 𝑧! in standard form given !
!
!
!
𝑧! = 25 2 cos βˆ’ ! + 𝑖 sin βˆ’ ! and 𝑧! = 14 cos ! + 𝑖 sin ! Let’s apply the formula: 𝑧! βˆ™ 𝑧! = π‘Ÿ! βˆ™ π‘Ÿ! [cos πœƒ! + πœƒ! + 𝑖 sin πœƒ! + πœƒ! ]. So we have !
!
!
!
Note: When adding radian 𝑧! βˆ™ 𝑧! = 25 2 βˆ™ 14 cos βˆ’ ! + ! + 𝑖 sin βˆ’ ! + ! measures, don’t forget to find common denominators: πœ‹
πœ‹
πœ‹ πœ‹
πœ‹ 3 πœ‹ 4
= 350 2 cos
+ 𝑖 sin
βˆ’ + = βˆ’ βˆ™ + βˆ™ 12
12
4 3
4 3 3 4
3πœ‹ 4πœ‹
=βˆ’
+ 12 12
And using our calculator to find these values: πœ‹
= 12
= 478.11 + 128.11𝑖 8 Ex. Express the quotient of 𝑧! and 𝑧! in standard form. Use the same values as those in the previous example. !
!
Let’s apply the formula !! = !! cos πœƒ! βˆ’ πœƒ! + 𝑖 sin πœƒ! βˆ’ πœƒ! . !
!
𝑧! 25 2
πœ‹ πœ‹
πœ‹ πœ‹
=
cos βˆ’ βˆ’
+ 𝑖 sin βˆ’ βˆ’
𝑧!
14
4 3
4 3
=
25 2
7πœ‹
7πœ‹
cos βˆ’
+ 𝑖 sin βˆ’
14
12
12
And using our calculator (in radian mode) to find these values: = βˆ’0.654 βˆ’ 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. !
Recall that the reciprocal of π‘₯ is !. 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 βˆ’πœƒ + 𝑖 sin βˆ’πœƒ ] 𝑧 π‘Ÿ
!
!
Ex. Find the reciprocal of 𝑧 = 4 cos ! + 𝑖 sin ! Using our formula for the reciprocal, 1 1
= cos βˆ’πœƒ + 𝑖 sin βˆ’πœƒ 𝑧 π‘Ÿ
1
πœ‹
πœ‹
= cos βˆ’ + 𝑖 sin βˆ’
4
3
3
9 !
If asked to write this number in standard form, then we can evaluate cos βˆ’ ! and !
sin βˆ’ ! to end up with =
1
πœ‹
πœ‹
cos βˆ’ + 𝑖 sin βˆ’
4
3
3
=
1 1
3
βˆ’
𝑖 4 2
2
=
10 1
3
βˆ’
𝑖 8
8