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Let $ L(z) $ be a primitive of $ 1/z $ in some region $ D\subset\mathbb{C}\setminus\left\{ 0\right\} $.

Prove that $ e^{L\left(z\right)}=cz $ for some $ c \in \mathbb{C} $.

Im not sure where to start, my initial thought was to use Lioville theorem:

if $ f $ is entire and there exists $A,B>0 $, and $ n\in \mathbb{N} $ such that $ |f(z)|\leq A+b|z|^n $ for any $z\in \mathbb{C} $ , then $ f $ is a polynomial with degree $\leq n $ in $ \mathbb{C}[z] $ .

But im not sure how to show that $ |f | $ indeed is bounded by such polynomial.

Any help would be appreciated, thanks in advance.

FreeZe
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1 Answers1

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Define $f(z) = z e^{-L(z)}$. Use $L'(z) = 1/z$ to show that $f'$ is zero, so that $f$ is constant.

We must assume that $D$ is connected (which might be implied by the term “region.”) Otherwise one can only conclude that $z e^{-L(z)}$ is constant on each connected component of $D$ (with possibly different values).

Liouville's theorem does not help here because that applies only to entire functions, i.e. functions which are holomorphic in all of $\Bbb C$.

Martin R
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  • Indeed, "region" means open and connected, I should have stated it. Thank you! – FreeZe Apr 13 '21 at 13:42
  • Can we use this result in order to prove that $L(zw)=L(z)+L(w)+a $ for some constant $ a\in \mathbb{C} $ ? – FreeZe Apr 13 '21 at 14:06
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    @FreeZe: The result implies that $e^{L(wz)-L(z)-L(w)}$ is constant. Alternatively show that the derivative of $L(zw)-L(z)-L(w)$ with respect to $z$ (for fixed $w$) is zero. – Martin R Apr 13 '21 at 14:11
  • The second way is clear. As for the first way, can we use regular power rules as wer do in $\mathbb{R} $? like $ e^{L(zw)-L\left(z\right)-L\left(w\right)}=\frac{e^{L\left(zw\right)}}{e^{L\left(z\right)}e^{L\left(w\right)}}=\frac{cwz}{cz\cdot cw}=\frac{1}{c} $ ? Also how the fact that it is a constant helps us? – FreeZe Apr 13 '21 at 14:24
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    @FreeZe: Yes, the same rules hold for the complex exponential function. – You'll need that if $e^{f(z)}$ is constant for a continuous function $f$ on a connected set, then $f$ is constant. – Martin R Apr 13 '21 at 14:33