# Homework Help: Linear algebra Homework question

1. Nov 1, 2009

1. The problem statement, all variables and given/known data
I am entirely lost with this one question I can't seem to figure out how to do it at all. The question states that $$\omega$$ is a complex number where $$\omega$$=(r)(e^(i $$\theta$$))
r and $$\theta$$ are real numbers
r>0
$$\theta$$ is element of [0,2$$\pi$$[
n is a positive integer

consider the equation z^n = $$\omega$$
Solve for z in terms of r, $$\theta$$ and n

2. Relevant equations

I would love to list some relevant equations but like I said, I have absolutely no idea what to do and any help at all would be greatly appreciated.

3. The attempt at a solution

I have tried looking over it multiple times however nothing seems to click. If anyone can just simply put me in the right direction or link me to anything that might help, that would be great!

-Nick

2. Nov 1, 2009

### tiny-tim

Welcome to PF!

(have a theta: θ and a pi: π and an omega: ω and try using the X2 tag just above the Reply box )

You're looking for a solution in the form z = se, such that zn = re

3. Nov 1, 2009

Hey :) Thanks for the welcome as well as the help.

I'm not too sure where to find all those symbols really :( New too all this!

and I guess that I replace z in the equation with z=se
so that Zn= [se]n = re

but then from there, how do you isolate each variable or constant so that you get some sort of answer?
sorry :( I'm just having a really hard time with this one...

Thanks again!

4. Nov 1, 2009

### tiny-tim

ok, and when you expand it, [se]n = … ?

5. Nov 1, 2009

well it ends up being

Zn=sn eiφn= re

but the path still seems blocked to me.... :(

6. Nov 2, 2009

### tiny-tim

(just got up :zzz: …)

ok, sneinφ= re

so now use the fact that the polar form, re, is unique in r, and unique in θ (mod 2π)

7. Nov 4, 2009

### rickz02

Sir Tiny-tim is my solution to the problem right?

zn = rei$$\theta$$
ln(zn) = ln(rei$$\theta$$)
nlnz = ln(rei$$\theta$$)
lnz = $$1/n$$ln(rei$$\theta$$)
lnz = ln(rei$$\theta$$)1/$$n$$
z = (rei$$\theta$$)1/n

8. Nov 5, 2009

### tiny-tim

(just got up :zzz: …)
erm … it's very long, and it isn't a solution.

You've started with zn = rei$$\theta$$, and ended with z = (rei$$\theta$$)1/n,

which is just a restatement of the question.

As I said, the solution has to be in the form z = se

you have to say what s is, and what φ is, in each solution.

9. Nov 5, 2009

### rickz02

But the problem is to solve for z, so I was thinking i just need to have z in one side and other parameters in the other side(Am I right???).

10. Nov 5, 2009

### tiny-tim

Yes, the LHS of your z = (re)1/n is excellent.

It's the RHS that isn't finished.

11. Nov 7, 2009

Hey! Sorry been busy for the past couple of days :(
Still stuck on this question though..
I'm slowly understanding how it works though :)

Soo Sneinφ=re
I'm supposed to solve for z, but does that mean that I have to solve for s as well as φ individually or not?... which are the modulus and the argument I believe?

and so z= (sneinφ)1/n how do I add in the
θ $$\in$$ [0,2π[ I know it has to go in there somewhere!!

Thanks again,
-Nick

12. Nov 8, 2009

### tiny-tim

Hey Nick!

(just got up :zzz: …)
Yes, that's correct

if sneinφ=re,

then that's exactly the same as saying sn = r and einφ = e
(have an ε )

Yes!

If einφ = e,

then that's exactly the same as saying that nφ = θ (mod 2π), or nφ = θ + a multiple of 2π.

13. Nov 8, 2009

ok okay thanks :) soo then

do I replace the φ with θ+2kπ? or rather nφ

and that's it? Or is there more to be done?

and shouldn't the answer in polar form be with φ as opposed to θ?

Thanks again :)

14. Nov 8, 2009

### tiny-tim

φ = … ?
Not following you.

15. Nov 8, 2009

I guess it's nφ = θ+2k so I substitute nφ for that.

My question is.. is that where it ends?

if the question states: put the solutions in the form z=se would we keep nφ or still substitute it for θ+2k?

I'm getting myself confused :S

Thanks!!

-Nick

16. Nov 8, 2009

### tiny-tim

Hi Nick!

(btw, you mean 2π, not 2)

Look, it's very simple …

you haven't done that yet!

φ = … ?​

17. Nov 8, 2009

Sorry I copy pasted it but apparently the pi didn't work :S

Okay so then the answer for what φ equals to is

φ= (θ+2kπ)/n

and sn=r so

s= r1/n

and that gives me the argument and the modulus
so now to solve for z I write

z= r1/nei(θ+2kπ)/n
= (rei(θ+2kπ))1/n

Thanks again btw!! :)
-Nick

18. Nov 9, 2009

Stop here!