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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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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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