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Section 6.5, Trigonometric Form of a Complex Number

Homework: 6.5 #1, 3, 5, 11-17 odds, 21, 31-37 odds, 45-57 odds, 71, 77, 87, 89, 91, 105, 107

1 Review of Complex Numbers

Complex numbers can be written as z = a + bi, where a and b are real numbers, and i =

-1. This

form, a + bi, is called the standard form of a complex number.

When graphing these, we can represent them on a coordinate plane called the complex plane. It

is a lot like the x-y-plane, but the horizontal axis represents the real coordinate of the number, and

the vertical axis represents the imaginary coordinate.

Examples

Graph each of the following numbers on the complex plane:

2 + 3i, -1 + 4i, -3 - 2i, 4, -i (Graph sketched in class)

The absolute value of a complex number is its distance from the origin. If z = a + bi, then

|z| = |a + bi| = a2 + b2

Example

Find | - 1 + 4i|.

| - 1 + 4i| =

1 + 16 = 17

2 Trigonometric Form of a Complex Number

The trigonometric form of a complex number z = a + bi is

z = r(cos + i sin ),

b . is called the argument of z. Normally,

where r = |a + bi| is the modulus of z, and tan = a

we will require 0 < 2.

Examples

1. Write the following complex numbers in trigonometric form:

(a) -4 + 4i

To write the number in trigonometric form, we need r and .

r=

16 + 16 = 32 = 4 2

4 = -1

tan = -4

= 3 ,

4

since we need an angle in quadrant II (we can see this by graphing the complex number).

Then,

-4 + 4i = 4

3 + i sin 3

2 cos 4 4

Note: You want to the leave the angle in your answer instead of simplifying. There are

several reason for this. First, we worked hard to get the angle. Second, it will be easier to

do certain mathematical operations if we have the angle, as we'll see later in this section.

(b) 2 - 2

3i

3

r = 4 + 12 = 48 = 4

3

9 9 3

tan = -2 3 = - 3

3?2 3

= 11 ,

6

since we need an angle in quadrant IV. Then, the trigonometric form is

4

3 cos 11 + i sin 11

36 6

2. Write the complex number 4(cos 4 + i sin 4 ) in standard form.

33

To go from trigonometric form to standard form, we only need to simplify:

4 cos 4 + i sin 4 = 4 - 1 - i 3 = -2 - 2 3 i

3 3 22

3 Products and Quotients of Two Complex Numbers

If z1 = r1(cos 1 + i sin 1) and z2 = r2(cos 2 + i sin 2) are two complex numbers in trigonometric

form, then

z1z2 = r1r2 cos(1 + 2) + i sin(1 + 2) and

z1 = r1 cos(1 - 2) + i sin(1 - 2)

z2 r2

Proof of Multiplication Formula

We use "FOIL" to multiply the two trigonometric forms, noting that i2 = -1:

z1z2 = r1r2(cos 1 + i sin 1)(cos 2 + i sin 2)

= r1r2(cos 1 cos 2 + i sin 2 cos 1 + i sin 1 cos 2 + i2 sin 1 sin 2)

= r1r2 cos 1 cos 2 - sin 1 sin 2 + i(sin 2 cos 1 + i sin 1 cos 2)

= r1r2 cos(1 + 2) + i sin(1 + 2) ,

where we used the sum formulas in Section 5.4 in the last line.

z1 ? z2.

To show that the division formula holds, you can use the multiplication formula and that z1 = z2

Examples

Carry out each of the following operations:

1. 3 cos + i sin ? 4 cos 7 + i sin 7

33 4 4

3 cos + i sin ? 4 cos 7 + i sin 7 = 3 ? 4 cos + 7 + i sin + 7

3 3 4 4 34 34

= 12 cos 25 + i sin 25

12 12

+ i sin ,

= 12 cos 12 12

since /12 is coterminal with 25/12.

2(cos 8 + i sin 8 )

2. 3 3

2 (cos + i sin )

22 2

2(cos 8 + i sin 8 ) 2 8 8

3 3 = 2 ? cos - + i sin -

2 (cos + i sin ) 2 3 2 3 2

22 2

= 2 cos 13 + i sin 13

66

= 2 cos + i sin ,

66

since 13/6 and /6 are coterminal angles.

4 DeMoivre's Theorem

DeMoivre's Theorem says that if n is a positive integer and z = r(cos + i sin ) is a complex

number, then

zn = rn(cos n + i sin n)

We can show this by using the multiplication formula from above n times.

Example

Calculate (3 + 3i)4.

Since this complex number 3 + 3i is not in trigonometric form, we need to first convert it to trigono-

metric form to use the above formula:

3 + 3i = 3

+ i sin

2 cos 4 4

Then,

(3 + 3i)4 = (3

+ i sin 4

2)4 cos 4 4 4

= 324(cos + i sin ) = -324

(Depending on the directions, you may not need to convert your answer to standard form, as we did

here.)

5 Roots of Complex Numbers

The complex number z = r(cos + i sin ) has exactly n distinct nth roots. They are:

n r cos + 2k + i sin + 2k ,

nn

where k = 0, 1, . . . , n - 1.

Examples

1. Find all square roots of i.

We can write i in trigonometric form as i = 1(cos + i sin ). Then, we use the formula with

22

r = 1, = , n = 2, and k = 0 and k = 1 to see that the two second roots of i are

2

2 + i sin 2 = cos + i sin = 2 + i 2

1 cos 2 2 4 4 2 2

+ 2 + 2 5 5

1 cos 2 + i sin 2 = cos + i sin = - 2 - i 2

2 2 4 4 22

How to convert sin to Cos? sine function can be changed to cosine and vice versa by adding 90 degrees and its multiples in domain of function so Sin (a+90)= cos a it is +ve as in angle lies in 2nd quad if a is less than 90 and sine is + ve in 2nd quad 15.4K views View upvotes Related Answer Awnon Bhowmik , studied at University of Dhaka

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