Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

1 Answer 1

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 $\Omega$.

share|improve this answer
    
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 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 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 at 20:13

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.