Complex polynomial theorem proof?

nihilismor
Posers and Puzzles 07 Oct '08 06:07
1. 07 Oct '08 06:07
Please prove that every complex polynomial function f(x) of degree n > or = 1 has at least one complex zero.

p.s Kudos to the cat that knows this without internet help.
2. fiftyonehz
Member 009
07 Oct '08 18:00
Suppose all zeros are real...

f(x) = sum_{i=1}^n (x-z_i), where z_i is in R

then all coefficients are real
3. 07 Oct '08 21:08
Originally posted by fiftyonehz
Suppose all zeros are real...

f(x) = sum_{i=1}^n (x-z_i), where z_i is in R

then all coefficients are real
I don't think that's what he's getting at - real numbers are complex.
4. 07 Oct '08 22:13
Originally posted by mtthw
I don't think that's what he's getting at - real numbers are complex.
Complex in this context means a number with a non-zero imaginary component, methinks.

I don't know if that includes numbers that are purely imaginary or not.

For the uninitiated, in its broadest sense, a complex number is any number of the form a + b*i, where a and b are any real number, and i is the square root of -1, a number that does not exist within the realm of real numbers.

a is call the real part and b*i the imaginary part.
5. 07 Oct '08 23:11
Yes, geepamoogle said what I meant when I meant what I said.