Holomorphic function on the unit disc

  • Thread starter iamqsqsqs
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  • #1
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Does there exist a holomorphic function f(z) on the unit disc and satisfies f(1/n) = f(-1/n) = 1/n^3 for every n in N?
 

Answers and Replies

  • #2
22,089
3,297
There does not even exist a continuous function that does this.
 
  • #3
9
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How can we vigorously prove that? I am thinking of construct a function g such that g(1/n) = g(-1/n) = 1/n^2 and consider f/g to do it. However I am stuck and cannot go on
 
  • #4
22,089
3,297
What would f(0) be?
 
  • #5
299
20
There does not even exist a continuous function that does this.

Last time I checked, [itex]z \mapsto |z|[/itex] was continuous...

iamqsqsqs, try looking at the zeros of f(z) - z^3. Do they form an isolated set of points?
 
  • #6
606
1
Last time I checked [itex]z \mapsto |z|[/itex] was continuous...


*** Last time I checked [itex]\,\,\displaystyle{\left|\frac{1}{n}\right|\neq \frac{1}{n^3}}[/itex] ...

DonAntonio ***



iamqsqsqs, try looking at the zeros of f(z) - z^3. Do they form an isolated set of points?

....
 
  • #7
299
20
Sorry, brain fart. I meant to say [itex]z \mapsto |z|^3[/itex]
 
  • #8
606
1
Sorry, brain fart. I meant to say [itex]z \mapsto |z|^3[/itex]


Hehe...yes, I supposed so. Happens to me all the time. Your answer to look at the zeroes of [itex]\,\,f(z)-z^3\,\,[/itex] pretty much wraps this up, though.

DonAntonio
 

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