I hope someone could enlighten me!
If $P(z)$ is a polynomial, shows that if $P(z)$ has no complex zero, then $\frac{1}{P(z)}$ is bounded.
I hope someone could enlighten me!
If $P(z)$ is a polynomial, shows that if $P(z)$ has no complex zero, then $\frac{1}{P(z)}$ is bounded.
"if $p(z)$ has no complex zero, then $\frac{1}{p(z)}$ is bounded"
Not really; the exponential function is one of the simplest counterexamples...
Amitesh Datta's outlined proof is absolutely correct. Let me give another approach, which is in some sense more direct (e.g. it is not a proof by contradiction). The differences between his answer (given first!) and mine are relatively minor.
We may assume that $P(z)$ is nonconstant.
Step 1: For any nonconstant polynomial function $P(z)$, $\lim_{z \rightarrow \infty} |P(z)| = \infty$. (This easily reduces to the case of $z$ a real variable, in which case it is familiar from calculus.)
Step 2: If for all $z \in \mathbb{C}$ $P(z) \neq 0$, then the function $f(z) = \frac{1}{P(z)}$ is a continuous [indeed holomorphic, but we won't need this] function from $\mathbb{C}$ to $\mathbb{C}$. Moreover, by Step 1, $\lim_{z \rightarrow \infty} f(z) = 0$.
Step 3: I claim that any continuous function $f: \mathbb{C} \rightarrow \mathbb{C}$ which vanishes at infinity -- i.e., has $\lim_{z \rightarrow \infty} f(z) = 0$ -- is bounded. Indeed, there exists $R_1 > 0$ such that for all $z$ with $|z| \geq R_1$, $|f(z)| \leq 1$. Moreover, since $\{z \in \mathbb{C} \ | \ |z| \leq R_1 \}$ is a closed disk in the complex plane, it is compact and thus every continuous function on $R_1$ is bounded: there exists $M \in \mathbb{R}$ such that for all $z$ with $|z| \leq R_1$, $|f(z)| \leq M$. It follows that $f$ is bounded on all of $\mathbb{C}$ by $\max \{M,1\}$.
Here is another way to construe Step 3: for any locally compact space $X$, one can give a meaning to a function $f: X \rightarrow \mathbb{C}$ vanishing at infinity: it means that for every $\epsilon > 0$, there exists a compact subset $K \subset X$ such that $|f(x)| \leq \epsilon$ for all $x \in X \setminus K$. But in fact this is equivalent to saying that $f$ extends continuously to the one-point compactification $X_{\infty} = X \cup \{\infty \}$ of $X$ with $f(\infty) = 0$. Then, since $f$ extends continuously to a function with a compact domain, it is bounded. Note that in this case the one-point compactification of $\mathbb{C}$ is nothing else than the Riemann sphere, and this way of looking at complex functions plays a big role in complex analysis.
Let us assume, for a contradiction, that the function defined by the rule $f(z)=\frac{1}{p(z)}$ is not bounded. In this case, there exists a complex number $z_n$ such that $\left|f(z_n)\right|>n$ for all positive integers $n$. Of course, this implies that $\left|p(z_n)\right|<\frac{1}{n}$ for all positive integers $n$.
Exercise 1: Prove that $\lim_{\left|z\right|\to\infty} \left|p(z)\right|=\infty$.
Exercise 2: Prove that the sequence $\{z_n\}_{n\in\mathbb{N}}$ is bounded. (Hint: use Exercise 1.)
Exercise 3: Prove that the sequence $\{z_n\}_{n\in\mathbb{N}}$ has a convergent subsequence. (Hint: use Exercise 2.)
Exercise 4: Prove that $z\to p(z)$ is a continuous function from $\mathbb{C}\to\mathbb{C}$.
Exercise 5: Deduce that $p$ has a complex zero and hence obtain a contradiction. (Hint: use Exercise 3 and Exercise 4.)
I hope this helps!