Let $f$ be an entire function with $|f(z)|\le 100\log|z|,\forall |z|\ge 2,f(i)=2i, \text{ Then} f(1)=?$
I am trying to solve this question using Cauchy's Inequality .
Now since given that function is entire therefore it is continuous on circle $|z|=2$ and have a maximum value of $|f(z)|$ in and on that circle , let's call it M that is $ M=max \{ |f(z)| : |z|≤2\}$.
Now for $|z|>2$ we have $$|f(z)|\le 100\log|z|<100|z|$$
Now consider that circle $|z-z_0|=R$ now on circle we have $$|z|≤|z_0+Re^{i\theta}|≤(|z_0|+R)$$
Thus if we define
$$M' = max\{M, 100(|z_0| +R)\}$$
Then we have $|f(z)|\leq M'$ Following by that $\lim_{R\to \infty} \frac{M'}{R^2}=0$ and Cauchy's Inequality we have
$$|f"(z_0)|\leq0 , \forall z_0$$
So using antiderivatives we have
I have doubt in this step , Is it right to integrate wrt z directly when f(z) is analytic ? )
$$f(z) = az +b$$ where a and b are complex constants .
Is this solution right or there is some mistake I made ?
I can get one relation from given condition , how to get other to eliminate a and b ?
