Square root of $e^{ix}$ Is this $$\sqrt{e^{ix}}=e^{\frac{ix}{2}}=\cos\frac{x}{2}+i\sin\frac{x}{2}$$ true?
Or
$$\sqrt{e^{ix}}=\sqrt{\cos x+i\sin x}$$
How do I express square root of $e^{ix}$ as a non-square root expression?
 A: They're both true. As well as $-\sqrt{\cos x+i\sin x}$ and $e^{ix/2+\pi}$.
While square roots are commonly avoided as much as possible when dealing with complex numbers, using natural exponents give $$\cos (nx)+i\sin(nx)=e^{inx}=(\cos x+i\sin x)^n$$ which is how I remember the double angle formula for sine and cosine, the sine and cosine of a sum of two angles, and so on.
A: Although "the" square root of $e^{ix}$ is not well-defined, every nonzero complex number $c$ has exactly two square roots in the sense that there are exactly two solutions of the equation $z^2=c$.
To find both square roots of $e^{ix}$ you must remember that $e^{ix}$ may also be written as 
$$e^{i x} = e^{i (x + 2 \pi n)} \quad\text{$n \in \mathbb{Z}$}
$$
The square roots of $e^{ix}$ all have the form
$$e^{i (x + 2 \pi n)/2} = e^{ix/2} \cdot e^{i\pi n} \quad\text{$n \in \mathbb{Z}$}
$$
Since $e^{i \pi n} = +1$ or $-1$ depending on whether $n$ is even or odd, we find that the two square roots of $e^{ix}$ are
$$e^{ix/2}, \quad -e^{ix/2}
$$
A: Assume that $x$ is real. You can use De Moivre's formula:
$$ \sqrt{e^{ix}} = e^{ix/2}, e^{ix/2+ i\pi}
$$
