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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$\begingroup$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 is thus zero, therefore the limit of $f'(z)$ to infinity is zero.
$\endgroup$ $\begingroup$That's wrong. $f(z) =z$ is analytic and has $$ f'(z) = 1 \to 1, \qquad z \to \infty $$
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