Is it true that $e^{i\omega}=1$ for any value of $\omega$? I have a problem understanding the polar form of a complex number. We have that
$$e^{i\omega} = \cos{\omega}+i\sin{\omega}.$$
In particular, if $\omega=2\pi$ then $e^{i2\pi} = \cos{2\pi}+i\sin{2\pi}=1$.
Yet, if I rewrite it like this: $$e^{i\omega} = e^{i2\pi\frac{\omega}{2\pi}},$$
then
$$e^{i\omega}=1^{\frac{\omega}{2\pi}}=1.$$
Does this mean that, for any value of $\omega$, $e^{i\omega}$ is always equal to 1?
 A: For complex numbers, we cannot say in general that $(e^z)^w=e^{zw}$. This stems from a problem of uniquely defining the power of one complex number to another. In other words, complex exponentiation is multi-valued.
To understand the multi-valuedness of complex exponentiation, sometimes I think it is helpful to distinguish between $\exp ix$, the natural exponential function with an imaginary exponent, and $e^{ix}$, the multi-valued imaginary power of a particular real number (which has the exponential function as its principal branch). This clarifies how we define the operation of raising a complex number to a complex power in general.
Defining complex exponentiation:
To start off with we can define the operation of raising a complex number to a natural number power very simply as repeated multiplication. So $z^2=zz$ and $z^3=zzz$ and so on. This gives us a unique answer for any $z^n$ with $z$ complex and $n$ natural. (This can be easily extended to integer powers as well.)
But to go further and define roots of complex numbers (so that we can define rational powers, so that we can define real powers by continuity), we immediately run into the problem that there are $n$ complex numbers $w$ satisfying $w^n=z$, and unlike the real number case, there is no natural choice for the "principal root" in general. Any choice (the root with the least angle from the positive real axis, for instance) is bound to create a discontinuity in the principal root function for a general complex input.
The natural exponential:
But we can define the natural exponential function for complex numbers, because it only needs natural number powers of complex numbers to work:
$$\exp z=\lim_{n\to \infty}\left(1+{z\over n}\right)^n$$
This gives a unique result for any input. The polar form of a complex number, and the Euler formula, are really referring to the value of the natural exponential function - that is, they only hold if you use the principal branch of $e^{ix}$.
Now to define the result of any complex number raised to the power of another complex number, we define one of the values of $e^{z}$ to be the natural exponential function, and then pretend the exponentiation identities hold. So if $z=r\exp i\theta$ and $w=x+iy$, then:
$$z^w=(r\exp i\theta)^{x+iy}=e^{(\ln r+i\theta)(x+iy)}=\exp(x\ln r -y\theta)\exp i(y\ln r+x\theta)$$
Multi-valuedness:
Why does this create a multi-valued function? Because it is also true that $z=r\exp i(\theta+2\pi k)$ for any integer $k$. So any of the following values are also valid results for $z^w$:
$$z^w=\exp(x\ln r -y(\theta+2\pi k))\exp i(y\ln r+x(\theta+2\pi k))$$
In general these values all lie on a logarithmic spiral (unless the exponent is real, in which case the possible values are all on a circle, or imaginary, in which case the possible values are all on a ray from the origin). Note that there are a finite number of unique values only in the case that the exponent is rational, and a single unique value only if the exponent is an integer.
This is how you can generate contradictions like $e^{ix}=1$ for all $x$ using complex exponentiation - by switching between the different possible results. So one possible result for $1^{x/2\pi}$ is $1$ like we would expect for a real exponential. This is from saying $1=1\exp i0$. But we can also write $1=1\exp i2\pi$, and get the result $1^{x/2\pi}=e^{ix}$.
