Patterns from complex numbers

[Matricaria]

Patterns from complex numbers ! URGENT!

- Use de moivre's theorem to obtain solutions for z^n=i for n=3, 4 and 5.
- Generalize and prove your results for z^n=a+bi, where |a+bi|=1.
- What happens when |a+bi|≠1


Relevant Equations:

z^n = r^n cis (n\theta)

r = \sqrt{a^2 + b^2}

\theta = tan^{-1}(\frac{b}{a})


I solved the first bullet:
Here's my solution for n=4 as an example:



z^4 - i = 0
==> z^4 = i
==> z = i^(1/4).

So this is equivalent to trying to find the 4 fourth roots of i.

In polar form, we see that:

i = cos(π/2) + i sin(π/2).

By De Movire's Theorem:

i^(1/4) = cos[(π/2 + 2πk)/4] + i sin[(π/2 + 2πk)/4]
==> i^(1/4) = cos(π/8 + πk/2) + i sin(π/8 + πk/2) for k = 0, 1, 2, and 3.

Thus:

k = 0 ==> z = cos(π/8) + i sin(π/8)
k = 1 ==> z = cos(5π/8) + i sin(5π/8)
k = 2 ==> z = cos(9π/8) + i sin(9π/8)
k = 3 ==> z = cos(13π/8) + i sin(13π/8)

I don't know how to apply generalization in this case..

 

[I like Serena]

Welcome to PF, Matricaria!

I prefer the form of De Moivre's theorem (or rather Euler's formula) that says:
$$a + i b = r e^{i\phi}$$
where $a = r \cos \phi$
and $b = r \sin \phi$Can you solve your equation $z^n = a+ i b$ with this form of the theorem?

It also effectively gives you the answer to the third question.If you are really required to use De Moivre's theorem as is, you can rewrite the exponential powers into cosines and sines.

 

[Matricaria]

Thank you very much for your help!
I was supposed to use De Moivre's as it is.. So I did use the sin and cos form..
Anyways, I have another question..

I can't find the modulus and argument of X^4=1?
I used De Moivre's Theorem to find the four roots which are: 1, -1, i, and -i..
I also plotted the argand diagram but can't calculate r and theta since all four points are on the axes...

 

[HallsofIvy]

Then you have serious problems understanding what the "Argand diagram" is! Both "r" and $\theta$, for 1, -1, i, and -i, should be trival simply by looking at the Argand diagram. Now "calculation" required!

 

[Matricaria]

We learned to calculate r using Pythagoras y'know after joining the lines into a right angle, and calculate theta using tan opp/adj!
Can you elaborate??

 

[I like Serena]

Thank you very much for your help!

You're welcome!

We learned to calculate r using Pythagoras y'know after joining the lines into a right angle, and calculate theta using tan opp/adj!
Can you elaborate??

Let's try and generalize this a bit.

If you have z = x + i y, this corresponds to a point with coordinates x and y.
You can draw your Pythagorean triangle using this point.
Now your "adj" is equal to x, and your "opp" is equal to y.

In other words:

$\tan \theta = {opp \over adj} = {y \over x}$,

and $r^2 = x^2 + y^2$.Now let's take for instance your solution "i".
What is x? What is y?
What would therefore r be?
And what would $\theta$ be?

 

[Matricaria]

x=0 and y=1

theta = 90 degrees?

 

[I like Serena]

x=0 and y=1

theta = 90 degrees?

Yep!

And r?

 

[Matricaria]

r = 1!

 

[I like Serena]

r = 1!

Right! :!)

So what about 1, -1, and -i?
Any thoughts?

 

[Matricaria]

z=-1 ---> x=-1 , y=0
r= 1 theta= 90 degrees as well

z=i ----> x=0 , y=1
r= 1

z=-i ----> x=0 , y=-1
r=1
!

What about theta when x=0?
Because tan1/0 is an error...

 

[I like Serena]

r=1
!

Yep!

z=-1 ---> x=-1 , y=0
r= 1 theta= 90 degrees as well

Hmm, that can't be right.
What is the tangent of 90 degrees?
And what is y/x in this case?

What about theta when x=0?
Because tan1/0 is an error...

So what's the inverse tangent of 0.9999/0.0001?



I recommend you take a look at the definition of the unit circle:
http://en.wikipedia.org/wiki/Unit_circle

I think it effectively answers your questions.

 

[Matricaria]

When z= -1, r= 1 and theta = 0??

And then z = i
x=0 and y=1
r=1 and theta= 90

And when z=-i
x=0 and y=-1
r=1 and theta = -90
?
Ok.. One last thing: I was supposed to come up with a conjecture after solvin' the three equations.. A conjecture for the distance/line segment, that is..
This is what I came up with: z^n = r*cis pi (n-2)/n
Is it any good?

Thank you very much for your help btw.. You're such a great tutor :)

 

[I like Serena]

When z= -1, r= 1 and theta = 0??

Almost.
theta = 180 degrees.

And then z = i
x=0 and y=1
r=1 and theta= 90

And when z=-i
x=0 and y=-1
r=1 and theta = -90
?

Yep!

Ok.. One last thing: I was supposed to come up with a conjecture after solvin' the three equations.. A conjecture for the distance/line segment, that is..
This is what I came up with: z^n = r*cis pi (n-2)/n
Is it any good?

Uhh...
I don't get it...
Which conjecture?

Thank you very much for your help btw.. You're such a great tutor :)

 

[Matricaria]

Use de moivre's theorem to obtain solutions for z^3-1=0
Use graphing software to plot these roots on an argand diagram as well as a unit circle with centre origin.
Choose a root and draw line segments from this root to the other two roots.
Measure these line segments and comment on your results.
Repeat the above for the quations z^4-1=0 and z^5-1=0. Comment on you results and try to formulate a conjecture.

 

[I like Serena]

Oh.
Didn't you in your original problem have z^n = i?
Is this the same problem?

Anyway, you seem to have skipped a step.
Did you measure line segments from one root to another?
Shouldn't your conjecture be about the length of such a line segment?

What do you mean by z^n = r*cis pi (n-2)/n then?
What is z supposed to be?
And what is r supposed to be?

 

[Matricaria]

Yeah, it's the same problems. I was workin' the roots for z^4=1 in the last line..

I drew all three equations (z^3=1, z^4=1, and z^5=1) on an argand diagram, and on a unit circle..
And I solved the three equations using De Moivre's and found 3, 4, and 5 roots for the three equations, respectively..
But, this is as far as I can go..

How do I measure the line segments?

 

[I like Serena]

Oh. Okay.

Well, measuring is usually something you do with a ruler...
But you should also be able to calculate them.
We'll get to that in a moment.

If you have plotted the solutions of z^5=1, did you see a pattern?
Can you describe how two neighboring roots are geometrically related?

 

[Matricaria]

I'll upload the three unit circles in a minute..

My observation was that any root for z^n=1 lies on the unit circle, and that the roots are equally spaced from one another around the circle

 

[Matricaria]

Those are the unit circles for the three equations

 

[Matricaria]

This is another one for the three equations on the same graph paper, just in case the other ones display a black background

 

[I like Serena]

Right!

Now, let's take a look at z^4 = 1.
Isn't that a square?
What are the lengths of its sides?

Can you calculate that?
Or can you perhaps measure it with a ruler?

 

[Matricaria]

This is the thing about this assignment, my teacher said we should do it analytically/algebraically, then find a pattern and put it into a conjecture !

 

[Matricaria]

And I don't know how to calculate the sides, that is!

 

[I like Serena]

You can draw a (isosceles) triangle between the origin and 2 neighboring roots.
What are the lengths of the sides of this triangle?
And what is the angle between the sides?

 

[Matricaria]

I think it's more of an equilateral triangle where each angle = 60 degrees?

And still don't know about the length of the sides :/

 

[I like Serena]

Hmm, if I look at the square, I intended the triangle between the points (0,0), (1,0), and (0,1).
That is definitely not an equilateral triangle...

Which triangle were you thinking of?

 

[Matricaria]

I was talkin' about the bigger one between (1,0) (-1/2, sqrt3/2) and (-1/2, -sqrt3/2)

About your traingle: It's a right angled traingle with angles: 90, 30 and 60..

Still no clue about the length

 

[I like Serena]

Ah, you were looking at z^3=1.
We'll get to that.

Okay, back to "my" triangle.
It is an isosceles triangle and its angles are actually 90, 45 and 45.
Have you ever heard of Pythagoras and how to calculate the sides of a right triangle?And let me also also try another tack here:
You have 2 points here: (1,0) and (0,1).
Do you know how to calculate the distance between 2 points?
Because that is $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$.

 

[Matricaria]

sqrt2?