Prove $\left(\frac{1+\cos2\theta+i\sin2\theta}{1+\cos2\theta-i\sin2\theta}\right)^n = \cos2n\theta+i\sin2n\theta$ 
Prove
  $$\left(\frac{1+\cos2\theta+i\sin2\theta}{1+\cos2\theta-i\sin2\theta}\right)^n = \cos2n\theta+i\sin2n\theta$$

Not sure how to go about this proof, was thinking of putting numerator and denominator in mod arg form straight away and then using binomial expansion but doesn't seem to work out... Then I tried realising the denominator first and ended up with
$$\frac{(1+cis2θ)^{2n} }{ 2^n(1+cos2θ)^n}$$  and not sure where to go from there?? Any help much appreciated, thanks :)
 A: We use Euler's formula : $e^{i\theta}=\cos\theta+i\sin\theta$
$$\left(\frac{1+\cos2\theta+i\sin2\theta}{1+\cos2\theta-i\sin2\theta}\right)^n $$
$$=\left(\frac{2\cos^2\theta+i(2\sin\theta\cos\theta)}{2\cos^2\theta-i(2\sin\theta\cos\theta)}\right)^n$$ $$=\left(\frac{2\cos\theta+2i\sin\theta}{2\cos\theta-2i\sin\theta}\right)^n $$ $$=\left(\frac{e^{i\theta}}{e^{-i\theta}}\right)^n $$ $$=e^{2in\theta}= \cos2n\theta+i\sin2n\theta$$.
A: The key point to note here is that the numerator and denominator have the same magnitude and opposite arguments. By drawing a unit circle centred on $1$, we can see that the argument of $1+\cos2\theta+i\sin2\theta$ is $+\theta$, so that of $1+\cos2\theta-i\sin2\theta$ is $-\theta$. Put together, we have
$$\frac{1+\cos2\theta+i\sin2\theta}{1+\cos2\theta-i\sin2\theta}=\cos2\theta+i\sin2\theta$$
The result follows by de Moivre's law.
A: $$\begin{align}
e^{i2\theta n} \cdot (1+e^{-i2\theta})^n
&=e^{i2\theta n}\cdot \left(\binom{n}{0} e^{-i2\theta \cdot 0} + \binom{n}{1} e^{-i2\theta \cdot 1}+\binom{n}{2} e^{-i2\theta \cdot 2} + \dots + \binom{n}{n-1} e^{-i2\theta \cdot (n-1)}+ \binom{n}{n} e^{-i2\theta \cdot n}\right) \\ 
&= \binom{n}{0} e^{i2\theta \cdot(n- 0)} + \binom{n}{1} e^{i2\theta \cdot (n- 1)}+\binom{n}{2} e^{i2\theta \cdot (n- 2)} + \dots + \binom{n}{n-1} e^{i2\theta \cdot (n-(n-1))}+ \binom{n}{n} e^{i2\theta \cdot(n- n)} \\ 
&= \binom{n}{0} e^{i2\theta \cdot n} + \binom{n}{1} e^{i2\theta \cdot (n- 1)}+\binom{n}{2} e^{i2\theta \cdot (n- 2)} + \dots + \binom{n}{n-1} e^{i2\theta \cdot 1}+ \binom{n}{n} e^{i2\theta \cdot 0}\\
&=\binom{n}{n} e^{i2\theta \cdot n} + \binom{n}{n-1} e^{i2\theta \cdot (n- 1)}+\binom{n}{n-2} e^{i2\theta \cdot (n- 2)} + \dots + \binom{n}{1} e^{i2\theta \cdot 1}+ \binom{n}{0} e^{i2\theta \cdot 0}\\
&=(1+e^{i2\theta})^n
\end{align}$$
Thus, 
$$
e^{i2\theta n}= \frac{
(1+e^{i2\theta})^n}{(1+e^{-i2\theta})^n} =  \left(\frac{
1+e^{i2\theta}}{1+e^{-i2\theta}} \right)^n=\left(\frac{
1+\cos2\theta +i \sin2\theta}{1+\cos2\theta -i \sin2\theta} \right)^n \ .$$
A: Here is a bit more general solution:
Let
$$z= e^{it}$$ 
Then, because of $\bar z = \frac 1z$, we get
$$w = \frac{1+z}{1+\bar z} = \frac{1+z}{1+\frac 1z}=z$$
Hence,
$$w^n = z^n = \cos nt + i\sin nt$$
