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Section 6.5, Trigonometric Form of a Complex Number - cos to sin calculator


Section 6.5, Trigonometric Form of a Complex Number-cos to sin calculator

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

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