I need to produce an example of a meromorphic function on $\mathbb{C}$ but not meromorphic on the Riemann sphere $\mathbb{C}_{\infty}$. Will this work: $f(z)=e^z-1/z$? Other examples are welcome. Thank you.
Example of a meromorphic function in $\mathbb{C}$ but not in $\mathbb{C}_{\infty}$
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complex-analysis
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1You probably mean $\mathcal C$ rather than $\mathbb C$; the latter denotes the complex numbers. – 2012-08-04
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3@MarcvanLeeuwen I dont understand really. – 2012-08-04
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1Katuk: @Marc seems to have thought that you were talking about the function spaces $\mathcal C$ and $\mathcal C^\infty$ rather then the complex plane and Riemann sphere. (He might have been been confused by the strange wording "meromorphic function in $\mathbb C$" rather than "on $\mathbb C$" or "function that is meromorphic on/in $\mathbb C$", since "in $\mathbb C$" modifies "meromorphic" rather than "function"). – 2012-08-04
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0Indeed I was put off by "in $\mathbb C$" since $\mathbb C$ contains no functions at all. I've now edited the question. – 2012-08-05
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I just wanted to note that your example $e^z-\frac1z$ works but so does the simpler $e^z$. (You can add in the $-\frac1z$ in order to put a pole in $\mathbb C$ if you like, but typically holomorphic functions are considered particular instances of meromorphic functions).
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0Yes, of course. But the most important feature of a meromorphic function is that it has poles(possibly none). So I don't think an entire function is a typical example of meromorphic functions. – 2012-08-05
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1Morgan's remark that we can take a holomorphic function is very clever precisely *because* it emphasises that although meromorphic functions immediately evoke poles, actually you don't *need* a function with poles to answer the question: the simplest (counter)examples are the best: +1 – 2012-08-05
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Meromorphic functions on $\mathbb{C}_{\infty}$ are rational functions. Let $f(z)$ be an entire function which is not polynomial, for example $e^z$ or sin $z$. Let $g(z)$ be an entire function which is not constant $0$. Suppose $f(z)/g(z)$ is not a rational function, for example let $g(z)$ be a polynomial. Then both $f(z)/g(z)$ and $g(z)/f(z)$ are meromorphic on $\mathbb{C}$ but not meromorphic on $\mathbb{C}_{\infty}$.
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0Ah! Thanks, I should have applied the result a function which is a meromorphic in the extended complex plane must be a rational function. – 2012-08-04