- #1
NewtonianAlch
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Homework Statement
Suppose c and (1 + ic)[itex]^{5}[/itex] are real, (c ≠ 0)
Show that either c = ± tan 36 or c = ± tan 72
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