Proof of Liouville's theorem - simple question

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Homework Statement


i'm using Fisher's Complex Variables for my complex analysis class and there is a proof for Liouville's theorem. It says "Set g(z)=(F(z) - F(0)) / z; Then g is an entire function"


Homework Equations


N/A


The Attempt at a Solution


I am confused by that statement. Doesn't g have a singularity at z=0? How can it be analytic over the entire complex plane?

Thank you and sorry if it seems like a very stupid question.
 
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I don't have that particular complex book. So, I am not sure about that particular proof. But do you think that for the simple reason that as z tends to 0 you will end up with the indiscriminate form 0/0 has something to do with it?
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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