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Complex Analysis - Manipulating trig identities

  1. Apr 10, 2012 #1
    1. The problem statement, all variables and given/known data
    Suppose c and (1 + ic)[itex]^{5}[/itex] are real, (c ≠ 0)

    Show that either c = ± tan 36 or c = ± tan 72

    3. The attempt at a solution

    So I considered the polar form [itex]\left( {{\rm e}^{i\theta}} \right) ^{5}[/itex] and that

    [itex]\theta=\arctan \left( c \right) [/itex], so c = tan θ

    Using binomial expansion, I expanded out the polar form exponential, and I consider only the imaginary part and equate that to zero, because it says the that (1 + ic)[itex]^{5}[/itex] is real.

    So that bit becomes:

    [itex]5\, \left( \cos \left( \theta \right) \right) ^{4}\sin \left( \theta
    \right) -10\, \left( \cos \left( \theta \right) \right) ^{2} \left(
    \sin \left( \theta \right) \right) ^{3}+ \left( \sin \left( \theta
    \right) \right) ^{5}[/itex]

    Now, I used MAPLE to check this and when I solve for θ, I get the values that I need to show that c = tan (x).

    How do I solve for θ by hand though? Also when I substitute cos^2 θ = 1 - sin^2 θ into Maple and then try solving that way, I get different values for theta, why is this? I thought doing the substitution might help simplify, but it changed the answer.

  2. jcsd
  3. Apr 10, 2012 #2


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    Hi NewtonianAlch! :smile:
    Since they give you the answer, try the obvious substitution …

    c = tanθ :wink:
  4. Apr 10, 2012 #3
    Uuuh, not sure why you converted to polar form...

    Can you just work out [itex](1+ic)^5[/itex] by working out the brackets?
  5. Apr 10, 2012 #4
    This time it worked...I just expanded it out, and then did the substitution, and solved for theta.

    I didn't go about it this way at first, because there was a first part to the question to prove that c = ±√5±2√5

    Doing it this way led me through the same path, and it asked to show a new method for the second part.
  6. Apr 11, 2012 #5


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    another method would be to simplify 1 + itanθ :wink:
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