The proof will be slightly different depending on whether the field is the real or the complex numbers. Here's how to do the real case.

If then . Write that as , which simplifies to . That is a quadratic in (with zero constant term), and if it is non-negative for all real then its discriminant cannot be positive. So , which obviously implies .

For the complex case, you have to feed a few complex conjugates into the calculation, but the method is essentially the same.