Does there exists a holomorphic function $f:B_2(0)\to\mathbb C$ such that $$f\left(\frac{1}{2n}\right)=f\left(\frac{1}{2n-1}\right)=\frac{1}{n},\;\forall n\in\mathbb Z^+.$$
I do not have an idea?
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1Think about the zero set of the function $f(z)-z/2$. – David C. Ullrich May 09 '16 at 17:42
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1@DavidC.Ullrich I think you mean $f(z)-2z$, no? – lulu May 09 '16 at 17:45
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@lulu Yes, thanks. – David C. Ullrich May 09 '16 at 19:36
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so, $0=f(z)-2z=f(z_n)-2z_n (n\to\infty)=f(\frac{1}{2n})-\frac{1}{n} (n\to\infty)= f(\frac{1}{2n-1})-\frac{2}{2n-1} (n\to\infty)=f(0)-0=0$ – serge May 09 '16 at 20:13
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Almost an exact duplicate of http://math.stackexchange.com/questions/299279 – mrf May 10 '16 at 07:46
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Let $$ z_n = \frac{1}{2n}, \qquad w_n = \frac{1}{2n - 1} $$ then $$ f(z_n) = 2z_n, \qquad f(w_n) = \frac{2}{(2n - 1) + 1} = \frac{2}{1/w_n + 1} = \frac{2w_n}{1 + w_n} $$ By the identity theorem we get the contradiction on the unit disc $$ 2z = f(z) = \frac{2z}{1 + z} $$
arney
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