Question

how do i find the roots of x^6-1 and x^12-1and then plot them on a complex plane?

Answer 1

sixth roots of one are ok. just find one of them.
\[x^6=1\]
\[x=1\] then divide the unit circle up into 6 equal parts

Answer 2

a picture is the easiest way. you will get 1, -1,
\[\frac{1}{2}+\frac{\sqrt{3}}{2}i\]

Answer 3

the conjugate
\[\frac{1}{2}-\frac{\sqrt{3}}{2}i\]

Answer 4

also
\[-\frac{1}{2}+\frac{\sqrt{3}}{2}i\]

Answer 5

and its conjugate

Answer 6

Okay, how did you get 1/2−3√/2i ? That seems to be where i am lost. Also, how do i split the unit circle into six compared to 4 or 8 sections of the unit circle

Answer 7

ok are you working in degrees or radians? doesn't matter just let me k now

Answer 8

u split a unit circle into 6 arcs of equal length

Answer 9

We are doing both but I would like to first work with degrees. It is easier for me to undertstand

Answer 10

see algebraically

\[x^{6} = 1 = e^{i2k \pi}\]

Answer 11

ok then you know that the circle has 360 degrees. divide this by 6 to get 60 degrees

Answer 12

so
\[x = e^{ik \pi/3}\]

Answer 13

yeah satellites right

Answer 14

so as you go around the circle you will go in multiples of 60 degrees:\0,60,120,180,240,300

Answer 15

since you are on the unit circle the first coordinate will be cosine and the second coordinate will be sine

Answer 16

cant we just do it algebraically and plot the points?

Answer 17

i should be more accurate and say "the real part will be cosine and the imaginary part will be sine"

Answer 18

@him yes of course but this is so easy if you have a picture of the unit circle in front of you. just divide into 6 parts and look at them

Answer 19

bt then the algebra will give her a better idea of why were plotting the points

Answer 20

I have the unit circle. How do i find cosine and sine

Answer 21

see dizliz

do u know how to plot e^(it)

Answer 22

?

Answer 23

look at the circle on the last page. the first coordinate is cosine and the second coordinate is sine

Answer 24

This is fairly straightforward in polar form....

Answer 25

first try to understand whether she knows wht the polar form means and how to plot a complex no

Answer 26

so, for example, at 60 degrees you see the point
\[(\frac{1}{2},\frac{\sqrt{3}}{2})\]

Answer 27

thinking of this as a point on the unit circle in the complex plane, the real part is
\[\frac{1}{2}\] and the imaginary part is
\[\frac{\sqrt{3}}{2}\]

Answer 28

\[as\ e^{i \pi/3} = \cos(\pi/3) + isin(\pi/3)\]

Answer 29

so if you look at the cheat sheet i sent, you will see at all the multiples of 60 degrees the coordinates of the point. those are your answers. do you see where i mean?

Answer 30

so the real part is cos (pi/3) and ther imaginary part is sin(pi/3)

Answer 31

@him, no. @satellite I understand what you are saying about imaginary and real but not exactly sure where those numbers come from. Is there a formula?

Answer 32

wht do u mean no? @dizliz
@sat: i told u weve gotta explain

Answer 33

see well show u

do u agree

\[1 = e^{i2k \pi}\]?

Answer 34

where k is an integer

Answer 35

are you asking where the numbers
\[\frac{1}{2}\] amd
\[\frac{\sqrt{3}}{2}\]come from?

Answer 36

yes

Answer 37

lets explain satellite

Answer 38

I know that a squared + b squared = y squared. Not sure if that helps any

Answer 39

yes and i this case y = 1 since you are on the unit circle

Answer 40

so you know that
\[a^2+b^2=1\]

Answer 41

yes that much I know

Answer 42

as to where these particular coordinates come from, we can find them from scratch if you like. but i am presuming that before you were asked to find the 6th roots of 1 you had some trig yes?

Answer 43

otherwise it is an unfair question. you should be familiar with basic trig before being asked to find roots of 1 in the complex plain

Answer 44

i agree

Answer 45

need to know the euler form and polar form

Answer 46

Right now I am taking a summer course. I feel like all i do is take notes and he is s o unorganized so it is really hard to read his notes. I am taking precal 2 and this is the first I have ever seen any of this

Answer 47

ok so lets start from scratch

Answer 48

do we get into it satelite?

Answer 49

he gave us euler's formula but with no explanation on how to use it or what it means

Answer 50

there are certain basic numbers (angles) for which you should know the values of sine and cosine

Answer 51

and in fact rather than derive these numbers which usually come from "reference triangles" let me ask you if by looking at the cheat sheet i sent, can you find
\[\sin(120)\]?

Answer 52

Assuming a knowledge of polar form (which is not hard to pick up if you have trig).

Take z = 3

For z^3 = 1, there are 3 roots of unity.

Represent z by <r, theta) and 1 by <1,0> so

z^3 = <r^3, 3*theta> = <1,0>

Taking the parts r^3 = 1 and 3*theta = 2m pi for some integer m.

So z = < 1, 2m pi/3 > and you find the 3 roots with m = 1,2,3.

To convert to Cartesian form

<1, 2m pi/3> = cos ( 2m pi/3) + i sin ( 2m pi/3).

This works generally not just for case = 3.

Answer 53

i think we are starting with basic trig yes?

Answer 54

were trying to explain the basics

Answer 55

yes, it root of 3/2

Answer 56

so now u know the trig ratios dizliz?

Answer 57

ok and how did you find it?

Answer 58

ratios?

Answer 59

i meant from the cheat sheet, how did you find it?

Answer 60

I think we must assume a knowledge of trig and the basics of complex numbers or this question is just inappropriate.

Answer 61

Because it is the imaginary part that i seen at the 120 degree

Answer 62

since shes only taking a summer course..we can excuse her

Answer 63

second coordinate on the cheat sheet yes?

Answer 64

yes second coordinate

Answer 65

and how about
\[\cos(300)\]?

Answer 66

1/2

Answer 67

got it

Answer 68

ok and you did this just by looking at the unit circle. now your question may be
"where did these numbers come from?" or it may be "how do i solve the problem i posted?"
which one would you like to address first?

Answer 69

ur amazing satellite

Answer 70

where do these come from

Answer 71

in other words without a cheat sheet how would i know they are there and what they are

Answer 72

uve gotta memorize em

Answer 73

ull just learn em over time

Answer 74

ok that's good. but be warned that for practical purposes you should memorize something. let me find it for you

Answer 75

do u get it dizliz>??

Answer 76

dizliz?

Answer 77

u there?>

Answer 78

yes, I am here. So I just need to memorize the unit circle

Answer 79

ok first of all do you know what a 30-60-90 triangle looks like?

Answer 80

no you only have to memorize two triangles really

Answer 81

Yes, that much I can gather

Answer 82

a 30 -60 -90 triangle and a 45-45-90 triangle

Answer 83

yes u need to memorize only some values

Answer 84

not sure what you mean by 45-45-90

Answer 85

lets do the easy one first. imagine you have a 45-45-90 triangle. since two sides are the same and it is a right triangle, if we set the hypotenuse = 1 (so it fits right inside the circle) you know that
\[a^2+a^2=1\]
\[2a^2=1\]
\[a^2=\frac{1}{2}\] and so
\[a=\sqrt{\frac{1}{2}}=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}\]

Answer 86

means two angles are 45 degrees and it is a right triangle.

Answer 87

now look at the coordinates of 45 degrees on the unit circle. you will see that both the first and second coordinate is
\[\frac{\sqrt{2}}{2}\]

Answer 88

visualize placing the triangle in the circle so that the hypotenuse is the radius. those are the lengths of the sides

Answer 89

Ok, I have a little bump where a goes from square root of 1/2 to square root of 2/2

Answer 90

In radians, first figure is cos, second is sin

Answer 91

that is the triangle on the upper right of the sheet i just sent

Answer 92

oh let me unbump you

Answer 93

\[\frac{1}{\sqrt{2}}=\frac{1}{\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}}{2}\]

Answer 94

this is called "rationalizing the denominator. they are the same numbers, just the form is different

Answer 95

Okay, that's pretty easy.