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We must show that there does not exist a holomorphic function $h(z)$ on the domain $\mathbb C - \{0 \}$ such that $\exp(h(z)) = z$ on the complex plane. Can we do this without using complex integration?

I know the proof of the fact that there exists no function whose exponential is the identity function $id(z) = z$ on the complex plane which uses complex analysis.

We proceed by assuming such a holomorphic $h(z)$ exists. This means that $e^{h(z)} = z$. Differentiating, we get $1 = e^{h(z)} h'(z) = z h'(z)$. This means that $1 = z h'(z)$, or $h'(z) = 1/z$.

Now we compute $\oint h'(z)$ around the countour $c(t) = e^{2 \pi i t}$ in two different ways:

  • $\oint h'(z) = h(1) - h(1) = 0$, by using FTC and that the countor starts and ends at the same point: $1$.
  • $\oint h'(z) = \oint 1/z = 2 \pi i$ by the Residue Theorem

Thus, we get $0 = 2 \pi i$ which is absurd.

However, this proof seems to rely on a lot of the machinery of complex integration to get things done. Is there no "simpler" proof? Can we show that to prove this fact, we somehow "need to" invoke facts about complex integrals?

Ideally, I would want an answer that only uses elementary properties of complex numbers, and properties of the complex exponential, and complex differentiation, but not complex integration.

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    $\begingroup$ What would $h(0)$ be? $\endgroup$ – Stephen Montgomery-Smith Aug 7 '20 at 22:04
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    $\begingroup$ You can prove it with the fundamental group which is very nice, but needs some basics in loops... $\endgroup$ – Mushu Nrek Aug 7 '20 at 22:09
  • $\begingroup$ I left it implicit that the domain of $h$ ought to be $\mathbb C - \{0 \}$ since this is what one "naturally expects". Edited for clarity $\endgroup$ – Siddharth Bhat Aug 7 '20 at 22:11
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    $\begingroup$ You can show that such an $h$ exists if the domain is restricted to $\mathbb{C}$ minus the nonnegative real axis (the various branches of the complex logarithm). Then show that none of the possible candidates for $h$ extend continuously to the positive real axis. $\endgroup$ – Qiaochu Yuan Aug 7 '20 at 22:19
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    $\begingroup$ Another simple way is to show by taking absolute values that the image of $h$ on any small punctured neighborhood of zero ($0<|z|<r<1$ will do) excludes the half plane $\Re z \ge 1$ say; this means that $h$ cannot have a singularity at zero (both poles and essential singularities have different local behaviour), so $h$ is continuable there to some finite value $a$ giving $e^a=0$ by analytic continuation; contradiction $\endgroup$ – Conrad Aug 7 '20 at 22:48
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Suppose such an $h(z)$ exists. Without loss of generality, we may select $h(1) = 0$ (the argument is similar for any other permissible choice of $h(1)$). Then for any $z, w \in \Omega := \Bbb{C} \setminus \{0 \}$, we must have $$e^{h(z)} e^{h(w)} = zw = e^{h(zw)},$$ which immediately implies the identity $h(zw) = h(z) + h(w) + 2\pi i k(z, w)$ for some continuous function $k: \Omega \times \Omega \rightarrow \Bbb{Z}$.

Since $\Omega \times \Omega$ is connected, $k(z, w)$ must be a constant $k$, and plugging in $z = w = 1$, we find $h(1) = 2h(1) + 2\pi i k$, so that in fact $k = -h(1)/2\pi i = 0$ and $h(zw) = h(z) + h(w)$ for all $z, w \in \Omega$.

By induction we get $$h(z_1 ... z_m) = h(z_1) + ... + h(z_m), \text{ for all } z_1, ..., z_k \in \Omega.$$

If we let $z_1 = z_2 = z_3 = -1$, then the above identity shows $h(-1) = 3h(-1)$, which is only possible if $h(-1) = 0$. But $e^0 = 1 \neq -1$, contradiction.

Edited to add: This type of approach often works well for showing that other inverse functions or multivalued functions (e.g. the square root $f(z) = \sqrt{z}$) cannot be extended holomorphically to $\Bbb{C}$ or $\Bbb{C} \setminus \{ 0 \}$. One can play with a functional equation or identity such a function must satisfy, and then derive a contradiction from the function needing to take on two different values at the same point (without having to use path or contour integration).

To show that the square root $f(z) = \sqrt{z}$ cannot be continuously extended to all of $\Bbb{C}$, for instance, one establishes that $f(zw) = \alpha(z, w) f(z)f(w)$, where $\alpha(z, w) \in \{ \pm 1 \}$; uses connectedness of $\Bbb{C}^2$ to establish that $\alpha(z, w)$ is constant for all $(z, w) \in \Bbb{C}^2$ (either $+1$ or $-1$); and then derives a contradiction from the fact that $\alpha(-1, -1)$ must have the opposite sign as $\alpha(+1, +1)$.(E.g. $f(1) = 1$, then $\alpha(1, 1) = 1$ but $\alpha(-1, -1) = -1$.)

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  • $\begingroup$ Your answer seems identical to the now-deleted older answer. What happened? (Note: Users with enough reputation points can always see deleted answers by default, and I am sure they will consider this situation quite awkward.) $\endgroup$ – Sangchul Lee Aug 8 '20 at 4:32
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    $\begingroup$ I’m quite aware of how “deleting” answers works, lol. Someone deciding to take down their answer, and then rethinking their decision, hardly seems like gripping, popcorn-worthy drama. I really hope people have better things to do with their lives, on a Friday night..... $\endgroup$ – Rivers McForge Aug 8 '20 at 4:55
  • $\begingroup$ I thought that your argument looked perfectly fine and even upvoted your answers, both original and this one, so I was simply curious. And of course I am a poor soul scrounging around this community at this wonderful Friday night, though that hardly sounds like something to be blamed... $\endgroup$ – Sangchul Lee Aug 8 '20 at 5:03
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    $\begingroup$ You know, I kind of felt like maybe having to use the topological concept of connectedness was too “high-powered of machinery” since the OP was asking us to use lower level stuff. But, obviously, they can decide for themselves how they feel about it. And I’m doing the same thing you’re doing right now, lol. Just killing time till a friend gets here, doodling some math questions.... $\endgroup$ – Rivers McForge Aug 8 '20 at 5:24
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    $\begingroup$ Actually, let me add on to my answer a little bit, I wanted to mention a generalization of this approach, and perhaps it will add interest for those super users who can read both of my answers. $\endgroup$ – Rivers McForge Aug 8 '20 at 5:54
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Let's answer the question in a more general context: There exists no continuous function $h:\mathbb{C}\setminus\{0\}\rightarrow\mathbb{C}$ such that $\exp \circ h = id\vert_{\mathbb{C}\setminus\{0\}}$. (Simply connected means that I can change every loop continuously into the null loop, i.e. a single point.)

For the proof, we need some concepts about loops and fundamental groups. Explain everything in detail would make this post into a script, so I will just give the ideas. If you draw a loop starting from (end thus ending in) some $z_0\in \mathbb{C}$, then you can continuously deform it into a single point. However, if you consider $\mathbb{C}\setminus\{0\}$, this is not the case anymore. Indeed, if a loop "goes around $0$, then it is impossible to continuously deform it into a point. By saying that two loops are equivalent iff we can continuously deform one into the other, we get an equivalence relation. In the example of $\mathbb{C}\setminus\{0\}$, two loops are equivalent iff they "go around $0$" the exact same number of times (in the same direction, i.e. clockwise or counter-clockwise). More generally, the equivalence classes of loops form a group with respect to the concatenation of loops. If we look at some (connected) open set $U\subseteq\mathbb{C}$ and some $z_0\in U$, then we write $\pi(U;z_0)$ for this group and we call it the fundamental group. This group describes "the holes" in $U$ in some sense. For example, $\pi(\mathbb{C}\setminus\{0\}, z_0) = \mathbb{Z}$, because we count, how many times a loop "goes around $0$" (positive for the counter-clockwise direction). [It is possible to define this notion in a much more general context!] As for any good structural description, we can translate it when we have a suitable transformation between to set $U,V\subseteq\mathbb{C}$. Indeed, if we have a continuous function $f:U\rightarrow V$, we may define $f(\alpha)$ for any loop $\alpha$ in $U$ as the image of the loop under $f$. Since $f$ is continuous, if $\alpha$ and $\beta$ are equivalent, then so are $f(\alpha)$ and $f(\beta)$. That means that we may define the function $\pi_f: \pi(U,z_0)\rightarrow\pi(V, f(z_0))$ through $\pi_f([\alpha]) := [f(\alpha)]$, where $[x]$ denotes the equivalence class of $x$. This is all we need, I hope it was not too confusing.

Suppose now there were such a function $h$. Since it is continuous, we may consider $\pi_h : \pi(\mathbb{C}\setminus\{0\}, z_0) \rightarrow \pi(\mathbb{C}, h(z_0))$. The same goes for $\pi_{\exp} : \pi(\mathbb{C}, h(z_0))\rightarrow \pi(\mathbb{C}\setminus\{0\}, z_0)$ and $\pi_{\exp\circ h} = \pi_{id_{\mathbb{C}\setminus \{0\}}} = id_{\mathbb{Z}} : \pi(\mathbb{C}\setminus\{0\}, z_0)\rightarrow \pi(\mathbb{C}\setminus\{0\}, z_0)$. It is easy to see that $$\pi_{\exp\circ h} = \pi_{\exp}\circ \pi_h$$. But since $\pi(\mathbb{C}, h(z_0)) = \{ 0\}$ (all loops can be continuously deformed to one point), we have $\pi_{\exp} = 0$ and thus $$ id_{\mathbb{Z}} = \pi_{\exp\circ h} = 0, $$ which is a contradiction.

You see that the proof is rather short if the basics are known. And it is worth while to take a closer look at fundamental groups. They are amazing objects and I wish I would have learned more about them before deviating to other domains... They initially came from Poincaré if I am not mistaken and they "characterise" pointed topological spaces. This website has some very nice and easy to follow explanations (and visualisations). Unfortunately, it is completely in French, but that is a beautiful language to learn anyway! (I am not French.)

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  • $\begingroup$ Thank you, the answer is very nice. I chose to accept the other answer because it uses less machinery than this one. I do agree that the fundamental group picture is very pretty ! $\endgroup$ – Siddharth Bhat Aug 8 '20 at 6:32
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There cannot be a continuous function $h$ on the unit circle $|z|=1$ such that $e^{h(z)}=z$. If there were, then a connectedness argument could be applied to prove that $h(e^{i\theta})=i\theta+2n\pi i$ for all $0 \le \theta < 2\pi$ and for some fixed integer $n$, which would contradict the assumed continuity of $h$ at $z=1$.

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