Turn on thread page Beta
    • Community Assistant
    • Thread Starter
    Offline

    17
    ReputationRep:
    Community Assistant
    So when calculating roots of complex numbers e.g. if z^n = r[cos(x) + isin(x)] then z = r^1/n[cos(x/n + 2pik/n) + isin(x/n + 2pik/n)] but why is it that you choose values up to k=n-1?

    Also why is it that in some problems you can do k=-1, k=-2, etc?
    Online

    22
    ReputationRep:
    (Original post by thekidwhogames)
    So when calculating roots of complex numbers e.g. if z^n = r[cos(x) + isin(x)] then z = r^1/n[cos(x/n + 2pik/n) + isin(x/n + 2pik/n)] but why is it that you choose values up to k=n-1?

    Also why is it that in some problems you can do k=-1, k=-2, etc?
    You know that the n^th roots of z, say \omega_i have to satisfy \omega_i^n = z. This is a polynomial equation of degree n, so can have at most n solutions.

    In your premise, taking k=0,1,2,3,... n-1 generates n different solutions (since cos and sine have least period 2pi) and so that must be all of the roots. To demonstrate this you have with k=0 the root z = r^{1/n}(\cos(x/n) + i \sin(x/n)) = r^{1/n}(\cos(x/n + 2\pi n / n) + i \sin(x/n + 2\pi n/n) which is the same as the root you get with k=0, since cos(y + 2pi) = cos y and sin(y+2pi) = sin y and 2pi * n/n = 2pi.

    Note that taking k=0,1,2,3,...n-1 is just convenient sometimes. In reality, taking k = {any sequence of n consecutive integers} will achieve the same thing, you will get n different roots and so that must be all the roots. In some problems, it is easier to take k = -n/2, -n/2 + 1,...,0,1,2,...,n/2 (for n even) but is equally valid.
    Online

    22
    ReputationRep:
    The above boils down to: i) there are at most n nth roots of a complex number (for n an integer) and ii) cos and sin have least period 2pi, so cos(y + 2pi k/n) differs from cos(y+2pi m/n) whenever k is not 0 modulo m.
    • Community Assistant
    • Thread Starter
    Offline

    17
    ReputationRep:
    Community Assistant
    (Original post by Zacken)
    You know that the n^th roots of z, say \omega_i have to satisfy \omega_i^n = z. This is a polynomial equation of degree n, so can have at most n solutions.

    In your premise, taking k=0,1,2,3,... n-1 generates n different solutions (since cos and sine have least period 2pi) and so that must be all of the roots. To demonstrate this you have with k=0 the root z = r^{1/n}(\cos(x/n) + i \sin(x/n)) = r^{1/n}(\cos(x/n + 2\pi n / n) + i \sin(x/n + 2\pi n/n) which is the same as the root you get with k=0, since cos(y + 2pi) = cos y and sin(y+2pi) = sin y and 2pi * n/n = 2pi.

    Note that taking k=0,1,2,3,...n-1 is just convenient sometimes. In reality, taking k = {any sequence of n consecutive integers} will achieve the same thing, you will get n different roots and so that must be all the roots. In some problems, it is easier to take k = -n/2, -n/2 + 1,...,0,1,2,...,n/2 (for n even) but is equally valid.
    Ahh, that makes more sense, thanks! So markers in the exam will give us the mark if we chose other ks that are convenient?
    Online

    22
    ReputationRep:
    (Original post by thekidwhogames)
    Ahh, that makes more sense, thanks! So markers in the exam will give us the mark if we chose other ks that are convenient?
    As long as your choice of k's work, yes.
    • Community Assistant
    • Thread Starter
    Offline

    17
    ReputationRep:
    Community Assistant
    (Original post by Zacken)
    As long as your choice of k's work, yes.
    Alright thanks!
    Online

    22
    ReputationRep:
    No problem
    Offline

    15
    ReputationRep:
    (Original post by thekidwhogames)
    Ahh, that makes more sense, thanks! So markers in the exam will give us the mark if we chose other ks that are convenient?
    You should take care to notice the allowed domain. Sometimes the question might specify a complex number with argument -\pi \leq \theta < \pi, so there you would need to choose solutions as appropriate to fit that domain. I tend to do it by finding roots 1, 2, 3, ... n-1 and then subtracting 2\pi from solutions out of the domain.
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: February 25, 2018
The home of Results and Clearing

2,613

people online now

1,567,000

students helped last year

University open days

  1. London Metropolitan University
    Undergraduate Open Day Undergraduate
    Sat, 18 Aug '18
  2. Edge Hill University
    All Faculties Undergraduate
    Sat, 18 Aug '18
  3. Bournemouth University
    Clearing Open Day Undergraduate
    Sat, 18 Aug '18
Poll
A-level students - how do you feel about your results?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.