(adsbygoogle = window.adsbygoogle || []).push({}); 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.

Thanks

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# Homework Help: Complex Analysis - Manipulating trig identities

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