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Let $ f $ be a non-constant entire function. Justify that $ f $ satisfies one of the following two statements:

(a) For all $ w \in \mathbb{C} $, the equation $ f(z) = w $ has a solution.

(b) For all $ w \in \mathbb{C} $, there exists a sequence $ (z_{n})_{n \in \mathbb{N}} $ of complex numbers such that $ |z_{n}| \to \infty $ and $ f(z_{n}) \to w $.

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You should rewrite point (b) ... there exists a sequence such that its image does what? is what? –  wisefool Jan 10 '13 at 0:28
    
You should first and foremost include your thoughts, what you tried and why you think this failed. –  Did Jan 10 '13 at 0:40
    
failed? i don't understand you. –  Rafael Jiménez Guerra Jan 10 '13 at 0:52
    
Are you posting this exercise because you don't know how to do it, right? Then, tell us what you did try, what you thought about the problem, how you planned to attack it and why all these eventually haven't worked. –  wisefool Jan 10 '13 at 0:58
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@Rafael: You are supposed to prove a single statement, which is that $f$ satisfies one of (a) or (b), so I do not know what it means to say that (a) is easy. What have you shown regarding (a)? –  Jonas Meyer Jan 10 '13 at 6:12
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1 Answer

up vote 1 down vote accepted

Let $ f $ be a non-constant entire function.


Case 1: $ f $ is a polynomial function.

By the Fundamental Theorem of Algebra, (a) is satisfied. Note that (b) cannot be satisfied because for any sequence $ (z_{n})_{n \in \mathbb{N}} $ in $ \mathbb{C} $, we necessarily have $$ \lim_{n \to \infty} |z_{n}| = \infty \quad \Longrightarrow \quad \lim_{n \to \infty} |f(z_{n})| = \infty. $$


Case 2: $ f $ is a transcendental function.

The Maclaurin series of $ f $ cannot terminate, otherwise $ f $ would be a polynomial function.

Consider the holomorphic function $ g: \mathbb{C} \setminus \{ 0 \} \to \mathbb{C} $ defined by $$ \forall z \in \mathbb{C} \setminus \{ 0 \}: \quad g(z) \stackrel{\text{def}}{=} f \left( \frac{1}{z} \right). $$ Note that $ g $ has an essential singularity at $ z = 0 $. This is because the left-tail of the Laurent series of $ g $ centered at $ z = 0 $ (obtained by making the substitution $ z \to \dfrac{1}{z} $ in the Maclaurin series of $ f $) does not terminate.

Applying the Weierstrass-Casorati Theorem, we see that for any neighborhood $ U $ of $ 0 $, the set $ g[U \setminus \{ 0 \}] $ is dense in $ \mathbb{C} $. Hence, for any $ w \in \mathbb{C} $, there exists a sequence $ (y_{n})_{n \in \mathbb{N}} $ in $ \mathbb{C} \setminus \{ 0 \} $ such that $ \displaystyle \lim_{n \to \infty} y_{n} = 0 $ and $ \displaystyle \lim_{n \to \infty} g(y_{n}) = w $. By setting $$ (z_{n})_{n \in \mathbb{N}} := \left( \frac{1}{y_{n}} \right)_{n \in \mathbb{N}}, $$ we obtain $ \displaystyle \lim_{n \to \infty} |z_{n}| = \infty $ and $ \displaystyle \lim_{n \to \infty} f(z_{n}) = w $. Therefore, (b) is satisfied.

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