Turn on thread page Beta

complex number question watch

Announcements
    • Thread Starter
    Offline

    0
    ReputationRep:
    hmm, a bit stumped here

    solve the equation: z^5 + 16 = 0


    anyone know how to do it? I think De Moivre's theorem is supposed to be used... rep to anyone who can do it
    Offline

    15
    ReputationRep:
    (Original post by kimoni)
    hmm, a bit stumped here

    solve the equation: z^5 + 16 = 0


    anyone know how to do it? I think De Moivre's theorem is supposed to be used... rep to anyone who can do it
    As

    (r (cos a + i sin a))^n = r^n (cos(na)+isin(na))

    we can see that if

    z^5 = -16

    we need z to have a modulus that is the fifth root of |-16| which is 2^(4/5)

    and will need one fifth of the argument which is pi (or 3pi or 5pi ...)

    So five roots are

    z = 2^(4/5) (cos(k pi/5) + i sin(k pi/5))

    for k = 1,3,5,7,9

    say.
    Offline

    2
    ReputationRep:
    (Original post by kimoni)
    hmm, a bit stumped here

    solve the equation: z^5 + 16 = 0


    anyone know how to do it? I think De Moivre's theorem is supposed to be used... rep to anyone who can do it
    You can do these sorts of things by thinking of circles and equally spaced points/lines on them to represent arguments. I get 16^(1/5).e(i.pi/5 + 2npi/5) ect. as solutions.

    Ben
    • Thread Starter
    Offline

    0
    ReputationRep:
    ok I think i'm beginning to see

    how did you know k was 1,3,5,7,9 though?
    Offline

    1
    ReputationRep:
    (Original post by kimoni)
    hmm, a bit stumped here

    solve the equation: z^5 + 16 = 0


    anyone know how to do it? I think De Moivre's theorem is supposed to be used... rep to anyone who can do it
    just curious, is this a question on its own or was there a prelim bit?
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by El Stevo)
    just curious, is this a question on its own or was there a prelim bit?
    just on it's own, but there are other qs like it
    Offline

    2
    ReputationRep:
    (Original post by kimoni)
    ok I think i'm beginning to see

    how did you know k was 1,3,5,7,9 though?
    Essentially, you've got the modulus and you're just trying to get the complex part to give you the -1 which you require to satisfy the equation. To get this, using e^i.theta, you can use theta = pi and, beacause of the periodicity of trig functions, you can use every angle 2pi larger. Then you just divide your angles by 5 to satisfy the question (as you take the 5th root, which is effectively dividing the exponential power by 5).

    Ben
    Offline

    15
    ReputationRep:
    (Original post by kimoni)
    ok I think i'm beginning to see

    how did you know k was 1,3,5,7,9 though?
    The argument of -16 is an odd multiple of pi.

    So the general answer is

    z = 2^(4/5) (cos a + i sin a)

    where a = k pi/5 and k is odd.

    But there aren't infinite roots, only five. So if you start counting through odd k then k=1,3,5,7,9 all give differents roots, but k=11 gives the same root as k=1, and k =13 the same as k=3, etc. This is because sin and cos have period 2pi.
    • Thread Starter
    Offline

    0
    ReputationRep:
    ok, thanks very much chaps
 
 
 
Turn on thread page Beta
Updated: January 8, 2005

University open days

  • University of East London
    Postgraduate Open Evening Postgraduate
    Wed, 23 Jan '19
  • University of East Anglia
    All Departments Open 13:00-17:00. Find out more about our diverse range of subject areas and career progression in the Arts & Humanities, Social Sciences, Medicine & Health Sciences, and the Sciences. Postgraduate
    Wed, 30 Jan '19
  • Aston University
    Postgraduate Open Day Postgraduate
    Wed, 30 Jan '19
Poll
Brexit: Given the chance now, would you vote leave or remain?
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

Equations

Best calculators for A level Maths

Tips on which model to get

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.