# Complex Analysis - Manipulating trig identities

1. Apr 10, 2012

### NewtonianAlch

1. The problem statement, all variables and given/known data
Suppose c and (1 + ic)$^{5}$ 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 $\left( {{\rm e}^{i\theta}} \right) ^{5}$ and that

$\theta=\arctan \left( c \right)$, 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)$^{5}$ is real.

So that bit becomes:

$5\, \left( \cos \left( \theta \right) \right) ^{4}\sin \left( \theta \right) -10\, \left( \cos \left( \theta \right) \right) ^{2} \left( \sin \left( \theta \right) \right) ^{3}+ \left( \sin \left( \theta \right) \right) ^{5}$

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

2. Apr 10, 2012

### tiny-tim

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

c = tanθ

3. Apr 10, 2012

### micromass

Staff Emeritus
Uuuh, not sure why you converted to polar form...

Can you just work out $(1+ic)^5$ by working out the brackets?

4. Apr 10, 2012

### NewtonianAlch

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.

5. Apr 11, 2012

### tiny-tim

another method would be to simplify 1 + itanθ