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.

  • $\begingroup$ It suffices to understand the formula $a+bi=re^{i\theta}$ where $r\cos(\theta)$ and $b=r\sin(\theta)$. Essentially you want to show that each complex number can be written in polar coordinates. $\endgroup$ Nov 15 '17 at 8:44
  • $\begingroup$ @Mathematician I haven't defined $\cos$ and $\sin$ yet. The plan is to define it via $\cos(z) := \frac{e^{iz}+e^{-iz}}{2}$ etc. $\endgroup$
    – blat
    Nov 15 '17 at 9:01

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.

  • $\begingroup$ For $\exp(\Bbb C)=z\exp(\Bbb C)$ we need $z\in \exp(\Bbb C)$, right? $\endgroup$
    – blat
    Nov 16 '17 at 19:28
  • $\begingroup$ @brot Yes, of course! Added the necessary detail. $\endgroup$
    – Pedro Tamaroff
    Nov 16 '17 at 19:51

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.

  • $\begingroup$ How do you know that such $r$ and $\vartheta$ exist? I think that's just a rephrasing of the problem. $\endgroup$
    – blat
    Nov 15 '17 at 18:39
  • 4
    $\begingroup$ @brot Because $r = e^x$ and I know $x$ exists? $\endgroup$
    – AmorFati
    Nov 15 '17 at 22:00
  • $\begingroup$ You wrote "For each $z \in \mathbb{C}$, we may write $z = r \exp(i \vartheta)$, for some $r >0$ and $\vartheta \in \mathbb{R}$". Can you show why we may do so? $\endgroup$
    – blat
    Nov 16 '17 at 6:48
  • $\begingroup$ @brot That was for context, to see the motivation $\endgroup$
    – AmorFati
    Nov 16 '17 at 8:42
  • $\begingroup$ Ok, can you prove that it's possible to write $z$ in that way? Your statement is $$\forall z\in \Bbb C \, \exists r\ >0 \,\exists \vartheta\ \in \Bbb R: z=r\exp(i\vartheta).$$ That's not trivial, so please show me why it is true. Keep in mind that I only have the series definition of $\exp$ and no Euler-formula or such things yet. $\endgroup$
    – blat
    Nov 16 '17 at 12:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.