$f$ be an entire function such that $f(0)=0, f(1)=2$ $f$  be an entire function such that $f(0)=0, f(1)=2$ Then
(i) there exist a sequence $(z_n )$ such that $|z_n | > n$ and $|f (z_n )| > n.$
(ii) there exist a sequence $(z_n )$ such that $|z_n | > n$ and $|f (z_n )| < n.$
(iii) there exist a bounded sequence $z_n  $ and $|f (z_n )| > n.$
(iv) there exist a sequence $(z_n )$ such that $|z_n | \to 0$ and $f (z_n ) \to 2 .$
$f(z)=z(z+1)$ is an entire and satisfis the condition,but this function does not satisfies (iv) by taking $z_n={1\over n}$, (i) is satisfied by taking $z_n=n+1$, but what  am I supposed to do here?  what about (ii),(iii)? thanks for the help.
 A: A most important step here is to understand the logic of the question/answers. If you say that (n) is true, you must demonstrate that it holds for all entire functions that satisfy the conditions $f(0)=0$ and $f(1)=2$.   To show that (n) is false, it suffices to give one counterexample: a function that is entire, satisfies $f(0)=0$ and $f(1)=2$, but fails (n).
(i) If you recall Liouville's theorem (a bounded entire function is constant), you should be able to show that (i) holds for every function $f$ satisfying the conditions. 
(ii) is   true for all entire functions that are not polynomials, but is false for polynomials of degree $2$ or higher. The reason is that for such polynomials $|f(z)|/|z| \to \infty$ as $|z|\to\infty$, which is incompatible with (ii). For the purposes of this exercise,  your counterexample $f(z)=z(z+1)$ is enough. 
(iii) and (iv) are false. You can disprove them with the same $f(z)=z(z+1)$ counterexample, or go further and show they are false for every $f$ that satisfies the conditions of the problem.
