Values of a complex polynomial

Solution to the following problem. Show that every non-constant polynomial p(x) will take every complex value, i.e. for any complex number c, there exists a complex number a such that p(a) = c.

The subtitle says it all - that if you take any polynomial (with complex number coefficients) of degree 1 or higher (i.e. constant "polynomials" are not allowed), then it will, at some point in the complex plane, take on any given complex value. (By complex numbers, we of course include the real numbers, which are merely a special type of complex numbers, as well.) The proof of this seemingly bold statement is surprisingly simple, given the Fundamental Theorem of Algebra.

To show that the statement is true, let us consider a related polynomial f(x), formed by subtracting the arbitrary constant c from the original polynomial p(x). That is, f(x) = p(x) - c.

But by the Fundamental Theorem of Algebra, f(x) necessarily has a number of complex number roots equal to its degree (the roots need not to be distinct). More importantly, f(x) has at least one root. That is, there exists at least one complex number a such that f(a) = 0 = p(a) - c, which of course means that p(a) = c. Since c is arbitrary, this says that at least one complex number a exists for which p(a) = c, regardless of the choice of complex number c. By letting the value of c runs through the entire range of complex numbers, we have the desired result that any non-constant polynomial with complex coefficients will take on every complex value somewhere. Simple enough, huh?

Note that the key point here is that we are working with complex numbers. The statement is completely absurd if we think in terms of real numbers only. For example, we all know (and can plainly see from its graph) that no matter what real number a we choose as the value for x, the polynomial p(x) = x² will never give a negative real value, let alone any number with an imaginary part. However, with the same example p(x) = x², and any complex number c, we can indeed find at least one (and usually two) complex value a such that p(a) = a² = c. For this particularly easy example, all we need to do in finding a is to use the quadratic formula on the polynomial f(x) = p(x) - c = x² - c, to get a = (plus or minus) sqrt(c). For instance, if c = -4, then a = 2i or -2i. If the number c has a nonzero imaginary part, then the expression for a can be a bit messy. But it can be simplified without too much difficulty. Say, if c = i, what value(s) of a will make p(a) = a² = i? The answers are of course the two square roots of i, which are equal to plus or minus [sqrt(2)/2 + i sqrt(2)/2]. Try it - square those two values to convince yourself that they are indeed the square roots of i. Another word, they are also two of the four complex fourth roots of -1!

This last bit give rise to another interesting question - how did I calculate the square roots of i? Well, that will be the topic for next issue.

Next time: What are the square roots of i?