Maths Help!

Discussion in 'Community Discussion' started by Shaun.P, Apr 17, 2006.

  1. Shaun.P macrumors 68000

    Shaun.P

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    #1
    Hi, I am in my first year of university and I have a class test tomorrow, worth 10% of my final grade. I have been looking at previous class test and a popular question seems to be complex roots. I can find any similar examples in my notes. Here is the question:

    Find in polar for the roots in C of the equation

    x ^ 5 = -32

    [4 marks]

    The test is out of thirty, so a question like this is worth more than a tenth of the paper.

    Any help you can provide is greatly appreciated. Am I right in saying there are 5 solutions, as it is x ^ 5?
     
  2. MongoTheGeek macrumors 68040

    MongoTheGeek

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    #2
    Yes,

    -2 is the obvious one.
    the full list
    2 cis (pi/5), 2 cis (3/5 pi), 2 cis (pi) (aka -2), 2 cis (7/5) pi, 2 cis (9/5) pi


    they are evenly spaced around the circle. The way you can figure it out is that when you multiple 2 complex number you add the angles. raising something to the 5 th power you multiply by 5. All 5 of the angles when multiple by 5 equal pi+n*2pi which is the same as pi.
     
  3. Shaun.P thread starter macrumors 68000

    Shaun.P

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    #3
    Hmm could you elaborate and tell me like I'm 5? :p

    I absolutely hate complex numbers, I have just found out how to do:

    x ^ 4 = 16i, but I don't understand how to do this question.

    What formula are you using? Could you go through it step by step please?

    Thanks again.
     
  4. Shaun.P thread starter macrumors 68000

    Shaun.P

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  5. Shaun.P thread starter macrumors 68000

    Shaun.P

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    #5
    A-ha! I have solved it I think. Is cis 'cos'?

    I drew an argand diagram, which helped me a lot.

    Thanks for your help.
     
  6. Shaun.P thread starter macrumors 68000

    Shaun.P

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    #6
    So, I put -32 in polar form, and I get:

    -32 (Cos[pi] + iSin[pi])

    Sin[pi] is 0, so, it is just:

    -32(Cos[pi])

    So it's the fifth root of -32(Cos[pi]) how do you proceed from here?

    I know it's something like 'fifth root of -32(Cos[[pi + 2k[pi]])' and you stick in values of k for 0, 1, 2, 3 and 4. But how do I get that -32 down to 2? :confused: :eek: :mad:
     
  7. mfacey macrumors 65816

    mfacey

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    #7
    Just out of interest: what are you studying?

    IT? Maths?
     
  8. Shaun.P thread starter macrumors 68000

    Shaun.P

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    #8
    Maths (Algebra and Calculus), Computer Science and Exploring the Cosmos (a really easy course because I couldn't understand physics). I intend to do a maths degree as I enjoy it and also want to do maths teaching at secondary / high school.

    This question is for a maths test I have. It's an algebra question. I prefer calculus.
     
  9. mfacey macrumors 65816

    mfacey

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    #9
    Not quite my cup of tea ;)

    Good luck on the test though!
     
  10. Shaun.P thread starter macrumors 68000

    Shaun.P

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  11. MongoTheGeek macrumors 68040

    MongoTheGeek

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    #11

    Cosine * i Sine. Its a way to express complex numbers as a vector.

    Take the standard X Y axis. Place the real numbers on the X axis and the imaginary numbers on the Y axis.

    There are 2 ways to express a complex number.

    One way is 4 +3 i. The other way is with a radius and an angle.

    4+3i is 5 * cos(37) + 5i*sin (37) or 5 cis 37

    As for formulae, I was never good at them and usually just re-derived every time.

    Lets take a look at the simple case of the square root of 4. Algebra tells us that the answers is +/- 2.

    In cis form it is 2 cis pi and 2 cis 0.

    4 is equal to 4 cis 0 , 4 cis 2 pi. 4 cis 4 pi, 4 cis 6 pi

    The square root absolute number 4 is 2.
    0/2 = 0
    2pi /2 = pi
    4pi /2 = 2 pi
    6pi /2 = 3 pi

    so the answers for square root of 4 is
    2 cis 0 = 2*cos 0 + 2i* sin 0 = 2+0i = 2
    2 cis pi = 2*cos pi + 2i* sin pi = -2+0i = -2
    2 cis 2 pi = 2*cos 2 pi + 2i* sin 2 pi = 2+0i = 2
    2 cis 3 pi = 2*cos 3 pi + 2i* sin 3 pi = -2+0i = -2

    The last 2 end up the same as the first 2 so there are only 2 interesting answers.


    In this case

    -32 = 32 cis pi (180 deg. the cosine of 180 is -1), 32 cis 3pi(540 deg or 180 deg + a full circle), 32 cis 5 pi , etc.

    The reason there are multiple angles is that when you can go around the circle any number of times and still end up at the same place.

    When you multiply two numbers in vector format you add the angles. So when you raise something to the 5th power you multiply the angle by 5, ergo when you take the 5th root you divide by 5.

    the 5th root of 32 cis pi would be 2 cis pi/5.

    There are other answers as well since -32 could also be 32 cis 3 pi that would have a different answer It would be 32 cis 3 pi/5.

    There are 5 answers before you reach 2 cis 13 pi/ 5 which is the same as 2 cis pi.
     
  12. Shaun.P thread starter macrumors 68000

    Shaun.P

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    #12
    I think I understand now. I have scanned my solution and uploaded to ImageShack. Can you take a look at it? I was taking my value of r to be -32, but I should of been taking the modulus of r, so it should be 32, right?

    Here.
     
  13. Shaun.P thread starter macrumors 68000

    Shaun.P

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    #13
    One final thing. The question asked 'In polar form' is my solution in polar form?


    Could you explain this:

    X^4 = 16i

    My tutor gets as the first solution:

    2 e ^ (i pi/8)

    And I get:

    2 (Cos pi/8 + i Sin pi/8)

    Are we both correct, is this just 2 different methods? The question wants the answer in polar form, are both these methods in polar form?

    Thanks again,
     
  14. MongoTheGeek macrumors 68040

    MongoTheGeek

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    #14
    Looks good to me.
     
  15. Shaun.P thread starter macrumors 68000

    Shaun.P

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    #15
    Thanks again for the help. I sat my test today and it had one of these questions in it for 3 marks!

    The other parts of the paper went OK but there were 2 differential equation questions where I didn't really know what I was doing!

    I'm hoping to get 22+/30.
     

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