Kernel of homomorphism $f(x)=e^{2\pi ix}$

Question

The book I am reading is giving an example of the first isomorphism theorem for groups. It says to consider the homomorphism $f:(\mathbb{R},+)\to(\mathbb{C}^{\times},\times)$ defined by $f(x)=e^{2\pi ix}$. It claims that the kernel of $f$ is $\mathbb{Z}$, but there's something that bothers me.

I thought: How come we don't just write $f(x)=e^{2\pi ix}$ as $f(x)=(e^{2\pi i})^{x}=1^{x}=1$? Then the kernel would be $\mathbb{R}$, wouldn't it?

Well, I do see that if $x=\frac{1}{2}$, for example, then $f(x)=e^{\pi i}=-1$, so not every real number is sent to 1. But I can't figure out what is making this inconsistency.

Am I misunderstanding the identity $e^{2\pi i}=1$? (The way I was taught that was by first writing out the series for cosine and sine.) Sorry if this is a really dumb question, I just couldn't figure out what was going wrong.

Answer 1

You didn't misunderstand the identity $e^{2\pi i} = 1.$ However, complex exponentiation is much more precarious than real-number exponentiation.

In particular, we define the complex exponential $$ z^w = e^{w \ln z} $$ where the complex natural logarithm of $z$ is multivalued, so in order to make sense of it, we need to choose a particular "branch" of the logarithm. In your case, when the appropriate branch is chosen, we have $$ \left(1\right)^x = e^{x \ln 1 } = e^{x \cdot 2\pi i} = e^{2\pi i x}$$ right back where we came from.

Answer 2

From Wikipedia, taking a real number $b$ and complex numbers $z,w$:

As, in general, $b^z$ is not a real number, an expression such as $(b^z)^w$ is not defined by the previous definition. It must be interpreted via the rules for powers of complex numbers, and, unless $z$ is real or $w$ is integer, does not generally equal $b^{zw}$, as one might expect.

Answer 3

For a complex number $z$, $z^{x}$ is not single valued and we cannot write $(e^{2 \pi i})^{x}=e^{2 \pi i x}$