# Roots of f(x)

1. Oct 15, 2005

### eljose

let be the function f(x) so we have that if x is a root also x* is a root, but we have that x is NEVER a pure imaginari number,i mean x is always different from x=ia the my question is if this means that all the roots will be real,the only counterexample i find is:

$$f(x)=(x-x_{0})(x-x_{1})(x-x_{2})........$$

that is an infinite polynomial that has all its roots in the form a is a root and also a*.

2. Oct 15, 2005

### Hurkyl

Staff Emeritus
(That's an infinite product, not a polynomial)

It is easy to construct counterexamples to the claim:

If f is a function such that
(1) whenever a is a root of f, then so is a*
(2) no root of f is purely imaginary
then the roots of f are real.

The simplest ideas work. We want f to have a non-imaginary non-real root. Let's pick (1+i).
Then, f must also have (1-i) as root.
So we need a function f that is zero at (1+i) and (1-i), and nonzero elsewhere.

That's very easy to do: let f be the function that is zero at (1+i) and (1-i) and 1 elsewere.

Last edited: Oct 15, 2005
3. Oct 15, 2005

### shmoe

Since I have a guess where you're trying to apply this, let's take the Riemann Zeta function as a counterexample. The roots come in conjugates by the reflection principle yet none of the roots are purely imaginary.

If you want something with no poles and not a polynomial, take f(x)=(x-1-i)*(x-1+i)*e^x.

4. Oct 15, 2005

### HallsofIvy

Staff Emeritus
Am I the only one who gets annoyed at the confusion of "roots" and "zeroes"?

My understanding is that equations have roots: 2 and 3 are roots of the equation x2- 5x+ 6= 0.
But that functions have zeroes: 2 and 3 are the zeroes of the function
x2- 5x+ 6.