I'm new to complex variable analysis. In the real case, $f:(\mathbb{R},+)\to(\mathbb{R^+},\cdot)$ defined by $f(x)=2^x$ is a homomorphism, furthermore, a isomorphism. In the complex case, I wonder if it can be generalized to explain something about complex exponentials.
I think there are a domain and a codomain of the function $z\mapsto2^z$ at the very beginning. However, I don't quite sure what is the domain and codomain here to be. Is it $f:(\mathbb{C},+,\cdot)\to(\mathbb{C},+,\cdot)$? And if so, it seems that $f(z)=2^z$ can hardly be a isomorphism. But being homomorphism is possible. So do we need to "shrink" the codomain so that injectivity holds? Next question is that, how to write down the way the function $f$ explicitly? For example, in polar transformation we can write something like $(r,\theta)\mapsto(r\cos\theta,r\sin\theta)$. Can $f(z)=2^z$ be written as so?
Update: My question is arised from the 3blue1brown video (I got lost at the time I mark). Besides knowing how can $z\mapsto 2^z$ be written in terms of $z$. I also wonder can $f(z)=2^z$ be written as componentwise formula? Namely, think of $\Bbb C$ as $\mathbb{R}^2$, writing the mapping as $(x,y)\mapsto\cdots\cdots$.
Yes, $f$ is a group homomorphism from $(\mathbb{C},+)$ onto $(\mathbb{C}\setminus\{0\},\cdot)$. However, it is not an isomorphism, since it is not injective: $f(0)=f\left(\frac{2\pi i}{\log 2}\right)=1$.
You can write $f$ explicitly:$$2^z=\exp(z\log 2).$$