Fundamental Theorem of Algebra. Every polynomial of degree greater than zero with complex coefficients has at least one zero.
Proof. Assume that \( p(z)=a_nz^n+a_{n-1}z^{n-1}+\cdots+a_0=0 \) has no solutions. Then \( 1/p(z) \) is entire, i.e. it is differentiable on \( \mathbb{C} \). Therefore \( 1/p(z) \) is also continuous. Also notes, as \( |z| \rightarrow \infty \), \( |1/p(z)| \rightarrow 0 \)
Thus,
\[ \exists M_R>0 :( \exists R>0:\forall z \in B_R=\{z\in \mathbb{C}: |z|\leq R\},
|1/p(z)| \leq M_R) \]
\( 1/p(z) \) is entire and bounded so it is a constant. But \( p(z) \) is not a constant. Contradiction, thus it must has at least one zero.
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