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Express $$ \left( \frac{1 + \sin\theta + i\cos\theta}{1 + \sin\theta - i\cos\theta} \right)^n $$ in the form $ x + iy$

$$ \left( \frac{1 + \sin\theta + i\cos\theta}{1 + \sin\theta - i\cos\theta} \right)^n = \left( \frac{1 + \sin\theta + i\cos\theta}{1 + \sin\theta - i\cos\theta} \cdot \frac{1 + \sin\theta + i\cos\theta}{1 + \sin\theta + i\cos\theta} \right)^n $$

$$ \left( \frac{\left( 1 + \sin\theta + i\cos\theta \right)^2}{\left(1 + \sin\theta\right)^2 - \left(i\cos\theta\right)^2} \right)^n = \left(\frac{\sin^2\theta}{1 + \sin\theta}+i\cos\theta\right)^n$$

then, if necessary, we can further expand via Newton's Binomial Theorem:

$$ \left( x + y \right)^n = {{n}\choose{k}}x^{n-k}y^k$$

Am I missing something?

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    Why you didn't use $\cos\theta+i\sin\theta$ instead of $\sin\theta+i\cos\theta$.? – 2017-01-20
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    @MyGlasses Maybe that's the way the problem was presented? But maybe it's a good idea to substitute $\alpha = \frac\pi2-\theta$ right away so that we have $\cos$ in the real part and $\sin$ in the imaginary part as we usually do. – 2017-01-20
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    So, with this assumption, you calculates are wrong: set $\dfrac{\pi}{2}$ in first and last. – 2017-01-20

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Go through exponential form.

$$\sin \theta+ i \cos\theta = i e^{-i\theta}$$ $$\sin \theta- i \cos\theta = -i e^{+i\theta}$$

Your expression is: $$\left(\frac{1+ie^{-i\theta}}{1-ie^{i\theta}}\right)^n=$$ $$\left(\frac{(i)e^{-i\theta}(-i)e^{i\theta}(1+ie^{-i\theta})}{1-ie^{i\theta}}\right)^n=$$ $$\left(\frac{i e^{-i\theta}(-ie^{i\theta}+1)}{1-ie^{i\theta}}\right)^n=$$ $$\left(i e^{-i\theta}\right)^n=\left( e^{-i\theta+i\pi/2}\right)^n=e^{in(\pi/2-\theta)}=\cos (n(\pi/2-\theta))+i\sin(n(\pi/2-\theta))$$

You'll notice the most awkward part is that sine and cosine are exchanged compared to the Euler's formula. For that reason, using a new variable $u=\frac{\pi}{2}-\theta$ for most of the calculation makes things much cleaner and show the true beauty of this solution:

$$\sin\theta+i\cos\theta=\cos u +i \sin u = e^{iu}$$ $$\sin\theta-i\cos\theta=\cos u -i \sin u = e^{-iu}$$ Now you don't have the awkward $i$'s there: $$\left(\frac{1+e^{iu}}{1+e^{-iu}}\right)^n=\left(e^{iu}\right)^n=e^{inu}=\cos nu+i\sin nu$$

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    That's $\cos (n(\pi/2-\theta))+i\sin(n(\pi/2-\theta))$, right? Yet another reason to substitute $u$. – 2017-01-20
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    Exactly. Thanks, fixed it. This substitution really saves everything. – 2017-01-20