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Complex Analysis - Value of imaginary part.

  1. Mar 10, 2012 #1
    1. The problem statement, all variables and given/known data

    Suppose both c and (1 + ic)[itex]^{5}[/itex] are real (c [itex]\neq[/itex] 0).
    Show that c = ± [itex]\sqrt{5 ± 2\sqrt{5}}[/itex]
    Now use another method to show that either c = ± tan 36◦ or c = ± tan 72◦


    3. The attempt at a solution

    I expanded it out, but I'm not entirely too sure how to solve this for c. Also, solving for c with wolfram gives c = i - which is not correct as c equals something else entirely.
     
  2. jcsd
  3. Mar 10, 2012 #2

    micromass

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    What did you get after expanding it out?
     
  4. Mar 10, 2012 #3
    After expanding and simplifying somewhat:

    1 + 5ic - 10c[itex]^{2}[/itex] - 10ic[itex]^{3}[/itex] + 5c[itex]^{4}[/itex] + ic[itex]^{5}[/itex]
     
  5. Mar 10, 2012 #4

    micromass

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    OK, that value is supposed to be real. So the terms with i should vanish. What does that give you?
     
  6. Mar 10, 2012 #5
    1 - 10c[itex]^{2}[/itex] + 5c[itex]^{4}[/itex]...

    Ah, so now I just factorise and solve?
     
  7. Mar 10, 2012 #6

    micromass

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    What do you mean by this?
     
  8. Mar 10, 2012 #7
    5c^4 - 10c^2 + 1 = 0
     
  9. Mar 10, 2012 #8

    micromass

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    Why should that be true?? Does that force (1+ic)5 to be real?
     
  10. Mar 10, 2012 #9
    I'm not too sure, if we solved for c this way, and substitute back into the original equation, then it should return a real value; i.e. no i component

    I just tried it in Maple and it returns a value with no i component.
     
  11. Mar 10, 2012 #10

    SammyS

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    You switched some coefficients around.

    That should be c4 - 10c2 + 5 = 0, if post #3 is correct.

    Completing the square should help with factoring.
     
  12. Mar 10, 2012 #11
    I can't see how those values switched around.

    In the post #3, c4 had a coefficient of 5.
     
  13. Mar 10, 2012 #12

    SammyS

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    Here are the terms with 1 .

    5ic - 10ic[itex]^{3}[/itex]+ ic[itex]^{5}[/itex]

    Factoring out ci gives:

    5 - 10c[itex]^{2}[/itex]+ c[itex]^{4}[/itex]
     
  14. Mar 10, 2012 #13
    I am really confused now. I thought that since we're only interested in the real terms here, any terms with the i component can just be cancelled out; hence leaving us with all the real terms, which is why I said

    5c^4 - 10c^2 + 1

    I'm not sure what you mean by here are the terms with 1.
     
  15. Mar 10, 2012 #14

    micromass

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    When is a+bi real??
     
  16. Mar 10, 2012 #15

    When b = 0.
     
  17. Mar 10, 2012 #16

    micromass

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    OK, so when is [tex]1 + 5ic - 10c^2 - 10ic^3 + 5c^4 + ic^5[/tex] real?
     
  18. Mar 10, 2012 #17
    I am tempted to say when c = 0, but c [itex]\neq[/itex] 0 according to the question.
     
  19. Mar 10, 2012 #18
    Oh hang on, I separated them out.

    When c^5 - 10c^3 +5c = 0?
     
  20. Mar 10, 2012 #19

    micromass

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    Note that

    [tex]1 + 5ic - 10c^2 - 10ic^3 + 5c^4 + ic^5=(1 - 10c^2 + 5c^4)+ i(5c - 10c^3 + c^5)[/tex]

    Do you see it now?
     
  21. Mar 10, 2012 #20

    micromass

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    Yes! So, you're given that [itex](1+ic)^5[/itex] is real. You've shown that this is equivalent to saying that

    [tex]c^5-10c^3+5c=0[/tex]

    Now you can factor this equation.
     
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