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Let $$\mu(z)=\frac{az+b}{cz+d}$$ be a Möbius map. Then one can show that $\mu$ is holomorphic on $\mathbb{C}\backslash\{\frac{-d}{c}\}$ since $\mu$ can be decomposed into four holomorphic functions. But if we consider $\mathbb{C}_\infty$ and use the definition of the derivative at $\frac{-d}{c}$ then we will see that $\mu'(-d/c)=\infty$.

Also, if $c=0$, then $\mu$ is obviously holomorphic on $\mathbb{C}$. But, again, if we check $\mu'(\infty)$ using the definition, we will see that it is $\frac{a}{d}$.

Why is this incorrect?

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    To be clear, what do you mean when you say you "use the definition of the derivative"? What computation are you doing? – 2017-02-10
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    $\mu'(z_0)=\lim\limits_{h\to 0} \frac{\mu(z+h)-\mu(z)}{h}$. But I guess I should use Euler-Lagrange equations. – 2017-02-10
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    The Möbius maps are holomorphic everywhere when viewed as functions from the Riemann sphere to itself (i.e. when you use an appropriate local parameter at infinity). Is that what you want to show? – 2017-02-10
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    @JyrkiLahtonen I want to show that it is holomorphic on $\mathbb{C}_\infty\backslash \{-d/c\}$. And if $c=0$ then it is holomorphic on $\mathbb{C}$. – 2017-02-10
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    Why do you think you have done anything incorrect? – 2017-02-10
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    What is $\Bbb{C}_\infty$? Presumably $\Bbb{C}\cup\{\infty\}$ (aka the Riemann sphre). What is the definition of a function being holomorphic at $\infty$? – 2017-02-10
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    @JyrkiLahtonen I think that if the limit above exists at infinity, then $\mu$ holomorphic at infinity. – 2017-02-10
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    I think not. If $z=\infty$ what is $z+h$? – 2017-02-10
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    @EricWofsey The thing is that the problem I'm working on asks to show holomorphism on those particular sets. – 2017-02-10
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    @JyrkiLahtonen I think we will have something like $\infty-\infty$. But in $\mathbb{C}_\infty$ this is probably not equal to $0$? And if $c=0$, we can take this limit with $\mu(\infty)$ to obtain $a/d$. – 2017-02-10
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    I would think that to define differentiability at infinity you need to use a different chart, and a different local variable. If this has not been explained in the accompanying material, then it is IMHO quite difficult to discuss differentiability at infinity. With proper charts you also get that $\mu$ is differentiable at $-d/c$. I was gambling on you studying a course on Riemann surfaces, and wanted to make you look up the definitions. If you are not, then I apologize. – 2017-02-10
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    @JyrkiLahtonen No, this is a complex analysis course. – 2017-02-10
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    People often define differentiable at infinity to mean $f(1/z)$ is differentiable at $0$. In this case, it is easy to check Mobius maps have that property. – 2017-02-10
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    What I was implying amounts to exactly what Alfred Yerger suggests. You do need to use definitions different from those of the more usual complex analysis. – 2017-02-10
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    @AlfredYerger Do you mean that $\mu$ is indeed holomorphic on $\mathbb{C}_\infty$? – 2017-02-10
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    I think it's likely that the OP has been studying calculus on the Riemann sphere without talking about charts. This is an annoying feature of some introductory complex analysis texts, so perhaps is missing the point of 'local variables' and such. Yes, it should be true that this map is holomorphic at $\infty$ and so everywhere on the sphere. Checking this amounts to checking that $\mu(1/z)$ is differentiable at $0$. So invert all the $z$ and then check that the result is differentiable at $0$. – 2017-02-10
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    @AlfredYerger The course is introductory, but the reference book we're using is of graduate level. I'll double-check in it about charts, which are from differential geometry, which I haven't studied yet. – 2017-02-10
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    The rational functions are the only holomorphic maps $\mathbb{C}_\infty \to \mathbb{C}_\infty$. $f(z)$ is holomorphic $\mathbb{C} \to \mathbb{C}_\infty$ iff for every $U \subset \mathbb{C}$ : $f(z)$ or $1/f(z)$ is holomorphic on $U$ – 2017-02-10

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