The analytic function at infinity

Question

Prove that if $ f(z) $ is analytic at the infinity, then: $$ \lim_{z\to\infty}f'(z)=0 $$

Answer 1

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

Answer 2

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.