# Show that there cannot be an entire function $F$ that satisfy the following condition

Show that there cannot be an entire function $F$ such that $F(x) = 1-\exp(2\pi i/x)$ for $1 \leq x \leq 2$. I think this has got something to do with Rouche's Theorem or the Argument Principle, but I'm not sure how to apply either of these to this specific problem. Can anyone shed some light?

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If $F(x)=1-\exp(2\pi i/x)$ for $1\leq x\leq 2$ and $f$ is entire, then in fact that equality holds over the whole of $\Omega=\mathbb C\setminus\{0\}$. Indeed, the two sides of the equality are holomorphic functions on the connected set $\Omega$ which coincide in a set which accumulates inside $\Omega$.

Now you should be able to check that there is no entire function which coincides with $1-\exp(2\pi i/x)$ on all of $\Omega$.

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Can you elaborate a bit more? I don't understand why the equality has to hold over the punctured complex plane. –  Aden Dong Nov 21 '12 at 18:54
See the «improvement» section in this Wikipedia page. This should be done in all complex analysis textbooks. –  Mariano Suárez-Alvarez Nov 21 '12 at 19:25
"...and $f$ is entire, then in fact that equality holds over the whole of $\Omega = \mathbb{C}\setminus {0}$." Why is this true? It is not obvious to me. –  Arturo don Juan Dec 10 '14 at 20:05
Oh ok it is obvious - the identity theorem. Woops. However, many textbooks don't cover this in the early section of power series. –  Arturo don Juan Dec 10 '14 at 20:12
Could you think of a proof/solution of this problem that doesn't involve the identity theorem? –  Arturo don Juan Dec 10 '14 at 20:13