Help solve $(1+\cos(a)+i \sin(a))^n$ I actually already have the solution to the following expression, yet it takes a long time for me to decipher the first operation provided in the answer. I understand all of the following except how to convert $\left(1+e^{i\theta \ }\right)^n=\left(e^{\frac{i\theta }{2}}\left(e^{\frac{-i\theta }{2}}+e^{\frac{i\theta }{2}}\right)\right)^n$
I am not sure if I wrote the expression correctly, I am new to this website.
Thank You!
 A: If you had $1+x^2$ you could write it as $x\left(\dfrac 1x+x\right)$ if you wanted to.
This occurs occasionally with the trigonometric/complex exponential functions because, of course $e^{-ia}=\dfrac 1{e^{ia}}$ and you can write the sine and cosine functions as a scalar multiple of something which looks like $x\pm\dfrac 1x$.
Of course what you have doesn't quite look like $1+x^2$ so you have to notice that form lurking beneath the surface. Worth noting the structure, though, as you will likely encounter it again.
And that is what is going on here - other answers will be likely more explicit. I just wanted to highlight this useful idea.
A: The first operation is just a result of algebraic manipulation. See here:
\begin{align} 
1 + e^{i\theta} = 
e^0 + e^{i\theta} = 
e^{i\frac{\theta}{2} - i\frac{\theta}{2}} + 
e^{i\frac{\theta}{2} + i\frac{\theta}{2}} =
e^{i\frac{\theta}{2}}e^{-i\frac{\theta}{2}} + 
e^{i\frac{\theta}{2}}e^{i\frac{\theta}{2}} = 
e^{i\frac{\theta}{2}}
\left(
e^{-i\frac{\theta}{2}} + 
e^{i\frac{\theta}{2}}
\right).
\end{align}
A: You can alternatively proceed as follows:
\begin{align*}
1 + \cos(x) + i\sin(x) & = 2\cos^{2}(x/2) +2i\sin(x/2)\cos(x/2)\\\\
& = 2\cos(x/2)(\cos(x/2) + i\sin(x/2))
\end{align*}
From this identity it results the desired claim:
\begin{align*}
(1 + \cos(x) + i\sin(x))^{n} = 2^{n}\cos^{n}(x/2)(\cos(nx/2) + i\sin(nx/2))
\end{align*}
Hopefully this helps!
