Proof the fundamental theorem of algebra
The fundamental theorem of algebra states that every non-constant polynomial with complex coefficients has at least one complex root.

Proof:
We will prove this theorem by contradiction. Suppose there exists a non-constant polynomial P(z) with complex coefficients that has no complex roots. Since P(z) is non-constant, it has a highest degree term, let's say az^n. We can assume that a is not equal to zero, otherwise, P(z) would be a constant polynomial.

Consider the function f(z) = 1/P(z). This function is well-defined for all z except for the roots of P(z). Since P(z) has no complex roots, f(z) is a continuous and bounded function on the complex plane. Therefore, by Liouville's theorem, f(z) is a constant function.

Let the constant value of f(z) be c. Then we have:

1/P(z) = c
P(z) = 1/c

Since P(z) is a non-constant polynomial, it has a highest degree term, let's say bz^m. We can assume that b is not equal to zero, otherwise, P(z) would be a constant polynomial. Substituting this into the equation above, we get:

bz^m = 1/c

But the left-hand side of this equation approaches infinity as z approaches infinity or negative infinity, which is a contradiction with the right-hand side being a non-zero complex number. Therefore, our assumption that P(z) has no complex roots must be false, and the fundamental theorem of algebra holds.

This completes the proof.
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Proof the fundamental theorem of algebra

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The fundamental theorem of algebra states that every non-constant polynomial with complex coefficients has at least one complex root.

Proof:
We will prove this theorem by contradiction. Suppose there exists a non-constant polynomial P(z) with complex coefficients that has no complex roots. Since P(z) is non-constant, it has a highest degree term, let's say az^n. We can assume that a is not equal to zero, otherwise, P(z) would be a constant polynomial.

Consider the function f(z) = 1/P(z). This function is well-defined for all z except for the roots of P(z). Since P(z) has no complex roots, f(z) is a continuous and bounded function on the complex plane. Therefore, by Liouville's theorem, f(z) is a constant function.

Let the constant value of f(z) be c. Then we have:

1/P(z) = c
P(z) = 1/c

Since P(z) is a non-constant polynomial, it has a highest degree term, let's say bz^m. We can assume that b is not equal to zero, otherwise, P(z) would be a constant polynomial. Substituting this into the equation above, we get:

bz^m = 1/c

But the left-hand side of this equation approaches infinity as z approaches infinity or negative infinity, which is a contradiction with the right-hand side being a non-zero complex number. Therefore, our assumption that P(z) has no complex roots must be false, and the fundamental theorem of algebra holds.

This completes the proof.
