Proving/Creating a conjecture on the roots of complex numbers

[Daaniyaal]

Homework Statement

Formulate a conjecture for the equation (z^3)-1=0, (z^4)-1=0 (z^5)-1=0
and prove it.

Homework Equations

r^n(cosnθ + isinnθ)

The Attempt at a Solution

Well my conjecture is that 2pi/n and 2pi/n + pi are possible values. I'm a bit iffy on how to word it. don't know which way I should go for a proof

 

[tiny-tim]

Hi Daaniyaal!

(try using the X² button just above the Reply box )

I'm a bit iffy on how to word it …

Start "if rⁿ(cosnθ + isinnθ) = 1, then …"

 

[Daaniyaal]

if rn(cosnθ + isinnθ)=1 then possible values for n are 2pi/n and 2pi/n+pi.

 

[tiny-tim]

i was thinking, more like …

if rⁿ(cosnθ + isinnθ) = 1,

then r = 1, and cosnθ = 1,

so nθ = … ? ​

 

[Daaniyaal]

0 most definitely

 

[Mark44]

0 most definitely

Aren't there other values for which cos(nθ) = 1?

 

[Daaniyaal]

2/3pi

 

[Daaniyaal]

By the way how would I put capital pi notation into this?

My conjecture is sqrt(2-2coskpi/n)) in pi notation: n-1, k=1

 

[Mark44]

Aren't there other values for which cos(nθ) = 1?

2/3pi

?
How did you get that?

 

[Mark44]

By the way how would I put capital pi notation into this?

My conjecture is sqrt(2-2coskpi/n)) in pi notation: n-1, k=1

What do you mean by "capital pi notation?"

 

[tiny-tim]

(just got up :zzz:)

cosnθ = 1,

so nθ = … ? ​

Come on, Daaniyaal …

what are all the solutions to cosψ = 1 ?​

(draw a graph of cos)

 

[Daaniyaal]

http://mathmaine.wordpress.com/2010/04/01/sigma-and-pi-notation/

2pi(n)

 

[Daaniyaal]

I think an alternate form of writing this would be e^(i x) = 1

 

[Daaniyaal]

and I can't get the superscript button to work :(

 

[tiny-tim]

2pi(n)

correct … ψ = 2π times n , for any value of n

but since we're already using n in the question, let's write that as …

ψ = 2π times k , for any value of k

sooo … what are all the solutions (for θ) of cosnθ = 1 ?

I think an alternate form of writing this would be e^(i x) = 1

yes

and I can't get the superscript button to work :(

do you have javascript turned off?

[noparse]alternatively, you can just type ^{before and} after[/noparse]

 

[Daaniyaal]

2,4,6,8,10,12 and so on?

 

[Daaniyaal]

All even numbers basically

 

[tiny-tim]

2,4,6,8,10,12 and so on?

θ = 2,4,6,8,10,12 … are the solutions to cosnθ = 1 ?

 

[Mark44]

http://mathmaine.wordpress.com/2010/04/01/sigma-and-pi-notation/

2pi(n)

This notation - $\Pi_{i =1}^n a_n$ - has nothing to do with what you're doing. It represents the product a₁*a₂* ... * a_n.

 

[Mark44]

2,4,6,8,10,12 and so on?

Daaniyaal, it would be helpful if you replied with complete statements. tiny-tim is trying to get you to do that with what he says below.

θ = 2,4,6,8,10,12 … are the solutions to cosnθ = 1 ?

 

[Daaniyaal]

$\Pi$$_{k=1}$$^{n-1}$

 

[Daaniyaal]

ahh fail

 

[Mark44]

$\Pi$$_{k=1}$$^{n-1}$

Again, this has nothing to do with what you're trying to do.

 

[tiny-tim]

$\Pi$$_{k=1}$$^{n-1}$

are you trying to write the set $\{\cdots\}$$_{k=0}^{n-1}$ ?

 

[Daaniyaal]

Yes, like in sigma notation

 

[Daaniyaal]

Okay. θ = 2pi,4pi,6pi,8pi,10pi,12pi if cosθ=1

 

[Daaniyaal]

Sorry about the short answers, I'm just so stressed with this project being due today, I will try to articulate properly.

 

[tiny-tim]

Okay. θ = 2pi,4pi,6pi,8pi,10pi,12pi if cosθ=1

let's write that as θ ε {2kπ}_k=0^∞

ok, now solve for λ if cos(nλ) = 1 (for a particular, fixed, n)

 

[Daaniyaal]

Okay. So if I go let n be 3

cos(3λ)=1
Then λ=(2 pi n)/3

 

[tiny-tim]

you mean λ ε {2 π k/3}_k=0^∞ ?

yes …

so what's the formula for a general n (instead of specifically n = 3)?

 

[Daaniyaal]

λ ε {2 π k/k}k=0∞

 

[tiny-tim]

where's n ?

 

[Daaniyaal]

where's n ?

Oh woops θ ε {2nkπ/n}k=0∞

 

[tiny-tim]

now, too many n's ! :rofl:

 

[Daaniyaal]

AAAAAAH. I swear there's a lot wrong up there. θ ε {2nkπ/k}k=0∞