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Prove that if $ f\mathrm{(}z\mathrm{)} $ is analytic at the infinity ,then : $ {\lim}_{{z}\mathrm{\rightarrow}\mathrm{\infty}}{f}\prime{\mathrm{(}}{z}{\mathrm{)}}\mathrm{{=}}{0} $

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Put $g(z) = f\left(\frac{1}{z}\right)$. Then $g(z)$ is analytic in a neighborhood of $z = 0$. This means that the derivative of $g(z)$ exists at $z =0$. We have

$$g'(z) = -\frac{1}{z^2}f'\left(\frac{1}{z}\right)$$

We can write this as

$$f'\left(\frac{1}{z}\right) =-z^2 g'(z) $$

The limit for $z$ to zero thus zero, therefore the limit of $f'(z)$ to infinity is zero.

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That's wrong. $f(z) =z$ is analytic and has $$ f'(z) = 1 \to 1, \qquad z \to \infty $$

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    $f(z) = z$ has a pole at infinity. – 2017-01-12
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    Yes sorry I have just edited it .its analytic at infinity . – 2017-01-12
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    Is there any difference ? – 2017-01-12
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    @user401187 Clearly $f(z) = z$ does not exist at infinity, so this function cannot be analytic there. From Liouville's theorem it follows that a function that is analytic everywhere and also at infinity, must be constant, the derivative of that is then zero. So, such cases are trivial. – 2017-01-12
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    The question was changed. – 2017-01-12