Patterns from complex numbers

[Matricaria]

Equals 1?

the distance of (cos2pi/4, sin2pi/4) and (1,0): sqrt2

the distance of (cos2pi/3, sin2pi/3) and (1,0): sqrt3

No pattern that I can spot :/ I'm sorry!

 

[I like Serena]

Errr... no, it does not equal 1.

You need the formula to calculate the distance between 2 points $(x_1,y_1)$ and $(x_2,y_2)$, which is:
$$distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$$

Do you understand how to use this formula?

 

[Matricaria]

I do!
I just did it with my calculator!

It's supposed to be Sqrt ((cos2pi/5)^2 + (sin2pi/5)^2)

It's sqrt1 :) :)

So, they's sqrt1, sqrt2, and sqrt3 for z^5, z^4, and z^3, respectively!
I can see a pattern: As the no of sides of the polygon increases by one, the distance between the roots decreases sqrt (n-1)

How can I formulate that into a conjecture??

 

[I like Serena]

Sorry to disappoint you, but it's not.
You've forgotten to subtract from 1.
It should be:
$$distance=\sqrt{(1 - \cos(2\pi/5))^2 + (0 - \sin(2\pi/5))^2}$$
Can you calculate that?

 

[Matricaria]

0.021932

 

[I like Serena]

Hmmm.
Does that look like a reasonable value?

Btw, do you know the difference between degrees and radians?

 

[Matricaria]

Yeah, I do!

Approx value 1.175570505?

 

[I like Serena]

Right!

So what would the formula be for n=4?
And for n=3?

Btw, that's it for me today.
I'm off to bed. :zzz:

 

[Matricaria]

I still can't formulate those findings into a conjecture I'm sorry...

Thank you very much for your help..
I woulda never asked for better help!

 

[I like Serena]

You've got this formula for n=5:
$$distance=\sqrt{(1 - \cos(2\pi/5))^2 + (0 - \sin(2\pi/5))^2}$$
What would this formula look like for n=4?

 

[Matricaria]

In case of n=4, we'll substitute the 5 for 4

and in case of n=3, we'll substitute the 5 for..

And for a general formula, we'll just put "n"...

Ok, one last thing, I was told that the conjecture would be in the form of z=rcis something, and I can't relate that to the distance

 

[I like Serena]

Okay.

We want to know the distance between the neighboring roots (cos2pi/5, sin2pi/5) and (1,0).
These points correspond to the complex values (cis2pi/5) resp. (1).
The vector between these 2 points is represented by $\Delta z = cis(2\pi/5) - 1$.
The length of this vector is $distance = |\Delta z| = |cis(2\pi/5) - 1|$.

 

[Matricaria]

So I can say that :

z^n= r*lcis(2pi/n) - 1l ?

 

[I like Serena]

Nope.

I'll repeat my original questions.
What does z or rather z^n represent here in your formula?
What does r represent?

I guess we could choose "r" to represent the distance, in which case we get:
$$r = |cis(2\pi/n) - 1|$$

 

[Matricaria]

z^n = lcis(2pi/n) - 1l*cis theta ?

and I guess theta would be equal to pi (n-2)/n (I guessed that from the rule for the interior angles of a polygon!)

 

[I like Serena]

Huh?

I really don't get what you intend with your formula.
Can you explain?

 

[Matricaria]

I mean, the original formula is: z=rcis theta
where r is the modulus and theta is the argument (For any complex number)..
YOu told me that we were going to substitute r for the distance between one root to another using the formula you helped me with..

So, I thought theta would be the interior angles of that polygon, ie the angle between each two sides..

 

[I like Serena]

No, I didn't intend for you to substitute r.

Your original formula defines the relation between any complex number z, and its distance to the origin r, and its angle theta.

However, your problem statement asks for the "distance" between different roots.
So as I see it, you need a formula that calculates a "distance" - not some complex number z.

What were you calculating?

 

[Matricaria]

This is the problem!
- 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.

Now all of it is done! I just need to formulate a conjecture which I can't!

 

[I like Serena]

Let's see what we can say.


z^n=1 has n roots given by z=cis(k 2pi/n), where k is the number 0, 1, ..., n-1.


The distance between 2 neighboring roots is |cis2pi/n - 1|.



However, I'm afraid this is not a conjecture, but just fact. ;)

 

[Matricaria]

Thank you very much!
I wouldn't ask for more :)

I'll work on the conjecture and let you know when I come up with something..

Thank you very much once again..
Have a great day!

 

[j.n]

i was just wondering if you found a conjecture for this? b/c i have to do this and i can't figure it out!
would it be |cis2pi/n - 1|?

thank-you so much!

 

[musicforlife]

Did you figure out the conjecture? i have NOO idea what to do

 

[musicforlife]

Hey. I am assuming you are done, and i happen to have the same topic now. Help with the conjecture , please ? :)