Find all $f$ such that $\vert f(z) \vert = \vert e^z \vert$

Asked Jul 12 '22 at 16:59

Active Jul 12 '22 at 17:07

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Let $f$ be holomorphic on $B_1(0)$ and continuous on $\overline{B_1(0)}$. Assume

$$\vert f(z) \vert = \vert e^z \vert$$

for all $z \in \partial B_1(0)$. Find all such $f$. Isn't this just going to be

$$f(n)=e^{\pi i n}$$

for every integer $n$. Or can I take

$$f_n(z):=\frac{n}{n+1}e^z?$$

edited Jul 12 '22 at 17:07

asked Jul 12 '22 at 16:59

homosapien

4,157

Are you claiming that $|e^z| = 1$ whenever $|z| = 1$? – Umberto P. Jul 12 '22 at 17:03
"Is the above equality equal to just $1$"? What does this mean – FShrike Jul 12 '22 at 17:03
What are you trying to deduce about the function anyway? – FShrike Jul 12 '22 at 17:04
@FShrike I thought that $\vert e^z \vert = 1$ on the unit circle? Im trying to find all such $f$ that have same modulus as $e^z$. – homosapien Jul 12 '22 at 17:05
$z=1$ is on the unit circle but $|e^1|=e\gt1$. Rethink... – FShrike Jul 12 '22 at 17:06
Aha good catch guys. I thought so too. – homosapien Jul 12 '22 at 17:06
2

No, $|e^z| = 1$ on the line $\text{Re}(z)=0$. – Robert Israel Jul 12 '22 at 17:07
@RobertIsrael thanks, brain fart there for a moment. – homosapien Jul 12 '22 at 17:07
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$f(n)$ isn't defined for $n>2$ since $f$ is defined on $B(0,1)$ – Didier Jul 12 '22 at 17:11
@Didier aha good catch! – homosapien Jul 12 '22 at 17:11
@Didier wait, are you referring to my $f(n)$ or my $f_n(z)$? – homosapien Jul 12 '22 at 17:13
What is $f(n)$ ? Aren't you looking for all functions that satisfy the inequality? Surely there are such functions which are defined at more than just integer values? – Carlyle Jul 12 '22 at 17:17
@Carlyle take a look at my $f_n(z)$. – homosapien Jul 12 '22 at 17:18
1

When $z = 1 + 0i$, $|e^z| = e$, but $f_1(1 + 0i) = \frac{e}{2} + 0i$ so $|f_1(z)| = \frac{e}{2} \neq e$ isn't it? – Carlyle Jul 12 '22 at 17:22
@HossienSahebjame: You are on the wrong track. Hint: Consider $g(z) = e^{-z} f(z)$. What can you say about $g$ on $\partial B_1(0)$? What can you conclude? – Martin R Jul 12 '22 at 17:22
@MartinR its modulus is bounded by $1$? on the boundary. – homosapien Jul 12 '22 at 17:33
@FShrike: No, not constant. – Martin R Jul 12 '22 at 18:33
@MartinR No? What’s my flaw – FShrike Jul 12 '22 at 18:51
@FShrike: What about $g(z) = z^7$? Or $g(z) = (z-2)/(1-2z)$? Look up “Blaschke factor”. – Martin R Jul 12 '22 at 18:52
@MartinR Excuse me, I misread the original post. I was thinking that $|f(z)|=|e^z|$ for all $z$, not just $z$ on the boundary – FShrike Jul 12 '22 at 18:53
Can we actually determine anything about $g$, from its having constant modulus on the circle? @MartinR I'm really curious what the strategy is – FShrike Jul 12 '22 at 19:05
Yes, @MartinR so am I – homosapien Jul 12 '22 at 19:07
1

See https://math.stackexchange.com/q/54386/42969 for a full characterization. – Martin R Jul 12 '22 at 19:08

Find all $f$ such that $\vert f(z) \vert = \vert e^z \vert$

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