Prove $\exp: \Bbb C \to \Bbb C^\times$ is surjective

Question

We define the complex exponential function: $$\begin{array}{rcl} \exp:\Bbb C &\to& \Bbb C^\times \\ z &\mapsto& \exp(z)=\displaystyle{\sum_{n=0}^\infty \frac{z^n}{n!}.} \end{array}$$

I wan't to show that this map is surjective. My idea is to show that the real exponential $\exp|_\Bbb R$ maps surjectively to $(0,\infty)$, then show that $\{\exp(ix)\,|\, x\in \Bbb R \} = \Bbb S^1$ and conclude $\exp(\Bbb C)=(0,\infty)\cdot\Bbb S^1 = \Bbb C^\times$.

This works because for any $z=a+ib \in \Bbb C$ we have $\exp(z)=\exp(a+ib)=\exp(a)\exp(ib)$. This is easy to show, using the definition and the cauchy product rule for series.

To show that $\text{Im}(\exp|_\Bbb R)=(0, \infty)$, we notice that $\exp(x)>1+x$ for $x>0$ and $\exp(0)=1$. So $\exp|_\Bbb R$ takes any value in $[1,\infty)$ by the intermediate value theorem. Since $\exp(-z)=\exp(z)^{-1}$ for any $z\in \Bbb C$, it also takes any value in $(0,1)$.

Unfortunately I don't know how to show the second part, i.e. $\{\exp(ix)\,|\, x\in \Bbb R \} = \Bbb S^1$.

Clarification: I don't have polar representation, trigonometry or any "advanced tools" yet. Just the power series definition of $\exp$ and basic analysis.

Answer 1

This is a general phenomenon that goes as follows. Suppose that $f : G\to H$ is a continuous group homomorphism with open connected image in $H$. Then $f$ is surjective.

Concretely, I will show that $\exp : \mathbb C\longrightarrow \mathbb C^\ast$ has open image and then note that a nonempty open subgroup of a connected topological group must be all of it.

To see $\exp$ has open image, note that we can define a function $\lambda : B(0,1)\longrightarrow \mathbb C$ such that $\exp\lambda(z)= 1+z$, so that $B(1,1)\subseteq \exp(\mathbb C)$. Explicitly, $\lambda(z)$ is defined as the usual powerseries for $\log(1+z)$ around the origin.

Since $\exp(\mathbb C) = z\exp(\mathbb C)$ for any nonzero complex $z\in \exp(\mathbb C)$, we see that $B(z,|z|)\subseteq \exp(\mathbb C)$ for any $z\in\exp(\mathbb C)$, so that indeed $\exp(\mathbb C)$ is open.

Now let us suppose $f : G\to H$ is as in the first paragraph, that is, $f$ is a continuous function $G\to H$ that is also a homomorphism of groups, and $f(G)$ is open and connected in $H$. Note that since $f$ is a homomorphism, $f(G)$ is a subgroup of $H$.

Let $A =f(G)$ denote and let $B$ be the complement $H\smallsetminus A$. Since $A$ is open, for each $h\in H$ the set $hA$ is open, and then $H = \bigcup_{h\in H} hA = A\cup B$ is open: recall that a group $H$ is always the disjoint union of cosets of any subgroup.

Then $B$ is also open and $H = A\cup B$. Since $H$ is connected and $A$ is nonempty, $B=\varnothing$ and $A=H$, which shows that $f$ is surjective, as we wanted.

Answer 2

Perhaps this might help.

For each $z \in \mathbb{C}$, we may write $z = r \exp(i \vartheta)$, for some $r >0$ and $\vartheta \in \mathbb{R}$.

With this in mind, if we instead write $z = x+iy$, we see that $$e^z = e^{x+iy} = e^x \cdot e^{iy},$$ where $x,y \in \mathbb{R}$. Notice that $e^x >0$ for all $x$ and $y \in \mathbb{R}$. By letting $r = e^x$ and $y = \vartheta$, we get the parallel between $e^z$ and the polar form for a nonzero complex number.

Let me know if you have any questions.