Proving that all entire & injective functions take the form $f = ax + b$?












2












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enter image description here



I'm a little confused at both the overall logic in this proof. Are we simply using $g(z)$ to make conclusions about $f(z)$, because $g(z)$ is the reciprocal of $f$? Is the proof assuming that $f$ is injective and entire (all the while knowing that it has some sort of singularity at $z = 0$), and then trying to reach contradictions in the essential singularity and removable singularity cases? Then, once it concludes that $z = 0$ is a pole singularity, it reaches the conclusion that $f$ must be of the form $f(z) = az + b$?



Also, more specific questions about the different cases:




  1. Removable singularity case: Why is $f$ bounded on the closed circle if $f$ is continuous? Am I missing something simple?


  2. Essential singularity case: Why exactly is $f({|z| > r } cap f({|z|<r}) neq emptyset$? $f({|z| > r })$ is dense, but how does $f({|z|<r}$ being open guarantee that their union is non-empty?







complex-analysis







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  • 1




    $begingroup$
    Can you give the reference to the picture you took? like the source of the problem?
    $endgroup$
    – BAYMAX
    Jul 12 '18 at 11:39
















2












$begingroup$


enter image description here



I'm a little confused at both the overall logic in this proof. Are we simply using $g(z)$ to make conclusions about $f(z)$, because $g(z)$ is the reciprocal of $f$? Is the proof assuming that $f$ is injective and entire (all the while knowing that it has some sort of singularity at $z = 0$), and then trying to reach contradictions in the essential singularity and removable singularity cases? Then, once it concludes that $z = 0$ is a pole singularity, it reaches the conclusion that $f$ must be of the form $f(z) = az + b$?



Also, more specific questions about the different cases:




  1. Removable singularity case: Why is $f$ bounded on the closed circle if $f$ is continuous? Am I missing something simple?


  2. Essential singularity case: Why exactly is $f({|z| > r } cap f({|z|<r}) neq emptyset$? $f({|z| > r })$ is dense, but how does $f({|z|<r}$ being open guarantee that their union is non-empty?







complex-analysis







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Can you give the reference to the picture you took? like the source of the problem?
    $endgroup$
    – BAYMAX
    Jul 12 '18 at 11:39














2












2








2


4



$begingroup$


enter image description here



I'm a little confused at both the overall logic in this proof. Are we simply using $g(z)$ to make conclusions about $f(z)$, because $g(z)$ is the reciprocal of $f$? Is the proof assuming that $f$ is injective and entire (all the while knowing that it has some sort of singularity at $z = 0$), and then trying to reach contradictions in the essential singularity and removable singularity cases? Then, once it concludes that $z = 0$ is a pole singularity, it reaches the conclusion that $f$ must be of the form $f(z) = az + b$?



Also, more specific questions about the different cases:




  1. Removable singularity case: Why is $f$ bounded on the closed circle if $f$ is continuous? Am I missing something simple?


  2. Essential singularity case: Why exactly is $f({|z| > r } cap f({|z|<r}) neq emptyset$? $f({|z| > r })$ is dense, but how does $f({|z|<r}$ being open guarantee that their union is non-empty?







complex-analysis







share|cite|improve this question









$endgroup$




enter image description here



I'm a little confused at both the overall logic in this proof. Are we simply using $g(z)$ to make conclusions about $f(z)$, because $g(z)$ is the reciprocal of $f$? Is the proof assuming that $f$ is injective and entire (all the while knowing that it has some sort of singularity at $z = 0$), and then trying to reach contradictions in the essential singularity and removable singularity cases? Then, once it concludes that $z = 0$ is a pole singularity, it reaches the conclusion that $f$ must be of the form $f(z) = az + b$?



Also, more specific questions about the different cases:




  1. Removable singularity case: Why is $f$ bounded on the closed circle if $f$ is continuous? Am I missing something simple?


  2. Essential singularity case: Why exactly is $f({|z| > r } cap f({|z|<r}) neq emptyset$? $f({|z| > r })$ is dense, but how does $f({|z|<r}$ being open guarantee that their union is non-empty?







complex-analysis




complex-analysis






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asked Oct 9 '14 at 0:21









Ryan YuRyan Yu

7731018




7731018








  • 1




    $begingroup$
    Can you give the reference to the picture you took? like the source of the problem?
    $endgroup$
    – BAYMAX
    Jul 12 '18 at 11:39














  • 1




    $begingroup$
    Can you give the reference to the picture you took? like the source of the problem?
    $endgroup$
    – BAYMAX
    Jul 12 '18 at 11:39








1




1




$begingroup$
Can you give the reference to the picture you took? like the source of the problem?
$endgroup$
– BAYMAX
Jul 12 '18 at 11:39




$begingroup$
Can you give the reference to the picture you took? like the source of the problem?
$endgroup$
– BAYMAX
Jul 12 '18 at 11:39










1 Answer
1






active

oldest

votes


















2












$begingroup$

The first part of what you say its right, thats what the author is trying to do.



For the other 2 specific questions.



1.$f$ Is bounded because the closed circle is a compact set and $f$ is continuous.



2.Because the set $f({|z|<r})$ is open it means that there is some ball inside $f({|z|<r})$ that only take points from $f({|z|<r})$, now because Cassorati-Weierstrass assures you that $f({|z| > r }$ is dense in all $mathbb{C}$, it must have some point inside that ball, thus the intersection, $f({|z| > r } cap f({|z|<r})$ is non-empty.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Sorry, why is the closed circle a compact set?
    $endgroup$
    – Ryan Yu
    Oct 9 '14 at 0:52






  • 1




    $begingroup$
    The Heine-Borel theorem tells us that in $mathbb{R}^n$, and thus in $mathbb{C}$, the compact sets are precisely those sets that are both closed and bounded.
    $endgroup$
    – JHance
    Oct 9 '14 at 1:19













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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









2












$begingroup$

The first part of what you say its right, thats what the author is trying to do.



For the other 2 specific questions.



1.$f$ Is bounded because the closed circle is a compact set and $f$ is continuous.



2.Because the set $f({|z|<r})$ is open it means that there is some ball inside $f({|z|<r})$ that only take points from $f({|z|<r})$, now because Cassorati-Weierstrass assures you that $f({|z| > r }$ is dense in all $mathbb{C}$, it must have some point inside that ball, thus the intersection, $f({|z| > r } cap f({|z|<r})$ is non-empty.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Sorry, why is the closed circle a compact set?
    $endgroup$
    – Ryan Yu
    Oct 9 '14 at 0:52






  • 1




    $begingroup$
    The Heine-Borel theorem tells us that in $mathbb{R}^n$, and thus in $mathbb{C}$, the compact sets are precisely those sets that are both closed and bounded.
    $endgroup$
    – JHance
    Oct 9 '14 at 1:19


















2












$begingroup$

The first part of what you say its right, thats what the author is trying to do.



For the other 2 specific questions.



1.$f$ Is bounded because the closed circle is a compact set and $f$ is continuous.



2.Because the set $f({|z|<r})$ is open it means that there is some ball inside $f({|z|<r})$ that only take points from $f({|z|<r})$, now because Cassorati-Weierstrass assures you that $f({|z| > r }$ is dense in all $mathbb{C}$, it must have some point inside that ball, thus the intersection, $f({|z| > r } cap f({|z|<r})$ is non-empty.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Sorry, why is the closed circle a compact set?
    $endgroup$
    – Ryan Yu
    Oct 9 '14 at 0:52






  • 1




    $begingroup$
    The Heine-Borel theorem tells us that in $mathbb{R}^n$, and thus in $mathbb{C}$, the compact sets are precisely those sets that are both closed and bounded.
    $endgroup$
    – JHance
    Oct 9 '14 at 1:19
















2












2








2





$begingroup$

The first part of what you say its right, thats what the author is trying to do.



For the other 2 specific questions.



1.$f$ Is bounded because the closed circle is a compact set and $f$ is continuous.



2.Because the set $f({|z|<r})$ is open it means that there is some ball inside $f({|z|<r})$ that only take points from $f({|z|<r})$, now because Cassorati-Weierstrass assures you that $f({|z| > r }$ is dense in all $mathbb{C}$, it must have some point inside that ball, thus the intersection, $f({|z| > r } cap f({|z|<r})$ is non-empty.






share|cite|improve this answer









$endgroup$



The first part of what you say its right, thats what the author is trying to do.



For the other 2 specific questions.



1.$f$ Is bounded because the closed circle is a compact set and $f$ is continuous.



2.Because the set $f({|z|<r})$ is open it means that there is some ball inside $f({|z|<r})$ that only take points from $f({|z|<r})$, now because Cassorati-Weierstrass assures you that $f({|z| > r }$ is dense in all $mathbb{C}$, it must have some point inside that ball, thus the intersection, $f({|z| > r } cap f({|z|<r})$ is non-empty.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Oct 9 '14 at 0:42









AramAram

1,4551022




1,4551022












  • $begingroup$
    Sorry, why is the closed circle a compact set?
    $endgroup$
    – Ryan Yu
    Oct 9 '14 at 0:52






  • 1




    $begingroup$
    The Heine-Borel theorem tells us that in $mathbb{R}^n$, and thus in $mathbb{C}$, the compact sets are precisely those sets that are both closed and bounded.
    $endgroup$
    – JHance
    Oct 9 '14 at 1:19




















  • $begingroup$
    Sorry, why is the closed circle a compact set?
    $endgroup$
    – Ryan Yu
    Oct 9 '14 at 0:52






  • 1




    $begingroup$
    The Heine-Borel theorem tells us that in $mathbb{R}^n$, and thus in $mathbb{C}$, the compact sets are precisely those sets that are both closed and bounded.
    $endgroup$
    – JHance
    Oct 9 '14 at 1:19


















$begingroup$
Sorry, why is the closed circle a compact set?
$endgroup$
– Ryan Yu
Oct 9 '14 at 0:52




$begingroup$
Sorry, why is the closed circle a compact set?
$endgroup$
– Ryan Yu
Oct 9 '14 at 0:52




1




1




$begingroup$
The Heine-Borel theorem tells us that in $mathbb{R}^n$, and thus in $mathbb{C}$, the compact sets are precisely those sets that are both closed and bounded.
$endgroup$
– JHance
Oct 9 '14 at 1:19






$begingroup$
The Heine-Borel theorem tells us that in $mathbb{R}^n$, and thus in $mathbb{C}$, the compact sets are precisely those sets that are both closed and bounded.
$endgroup$
– JHance
Oct 9 '14 at 1:19




















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