 We had to invent complex numbers because some equations with real coefficients didn't have real solutions. But this means we can write equations with complex coefficients. So the question you've got to ask yourself is, do we have to invent hyper-complex numbers to deal with these equations? Unfortunately, the answer is... Now, we're going to claim the following. We know about polynomial arithmetic for real polynomials is true for polynomial arithmetic of complex numbers. In other words, we don't have to worry about the details of adding, subtracting, multiplying, or dividing polynomials just because they happen to have complex coefficients. So let's start with the following claim. Let F be a degree N polynomial with complex coefficients. Then F, F conjugate will be a degree 2N polynomial with real coefficients only. Now, there's actually an easy way to prove this. But you don't train for a marathon by driving to work. And in fact, since we're talking about the fundamental theorem of algebra, let's talk about Gauss's strategy. Prove a theorem any way you can, then find another proof. And in fact, Gauss would find four different proofs of the fundamental theorem of algebra. So let's not take the easy way. Instead, well, let's try to prove a single case and see if we can generalize it. So we'll start by proving it's true for linear functions. So suppose we have a linear function az plus b, then the conjugate will be... and our product will be... Now, a, a conjugate and b, b conjugate are necessarily real. To prove that a, b conjugate plus b, a conjugate is also real, remember that a real number is its own conjugate. So we check. And so a, b conjugate plus b, a conjugate is also real. And so we know the statement is true for polynomials of degree N equal to 1. And we should probably do something at this point, but I'm not entirely sure what. Well, we'll leave it for you to figure out. So now let's consider f of z, a degree N polynomial with complex coefficients. The product f, f conjugate will be a polynomial with real coefficients. So its roots will be real and complex numbers. So the roots of f must also be real and complex, and we're done. Or are we? So we need to show that the real and complex roots of f, f conjugate include all the roots of f. So let's consider any root zi. z minus ci must be a factor of either f of z or f conjugate of z. Suppose z minus ci is not a factor of f of z. The division algorithm will allow us to write f of z as... where our i is a complex constant not equal to zero. Now, if z minus ci is not a factor of the conjugate of f of z, then we have where our i prime is another non-zero complex constant. And consequently, their product will be... and since z minus ci is a root of their product, we have... but this is impossible. And so z minus ci must be a factor of one of the polynomials. Now, since the conjugate is of degree n, only n of the roots can be roots of the conjugate. So the remaining roots must be roots of our original polynomial. And this proves that a degree n polynomial with real or complex coefficients has n roots among the complex numbers. And this gives us the real importance of the fundamental theorem of algebra. The real importance of the fundamental theorem of algebra is not that the roots of a polynomial can be found among the complex numbers. Rather, it's that there are no higher orders of numbers. Any polynomial equation you could write using complex numbers has all of its solutions among the complex numbers. We say that the complex numbers are algebraically closed, and there's no need to invent a new type of hypercomplex number.