Zeros of complex functions - simple question I know this is easy one but I can't get to the bottom of it. A zero of a real function is when its value is equal to zero. But what is a complex zero? When the real part of its value is equal to zero OR when its imaginary part is equal to zero OR both must be zero?
Thanks in advance.
 A: I was reluctant to post an answer, but I guess I can give more details than I did in my comment.
We have to remember that the complex number system has two parts - the real and imaginary parts. Whenever we write the symbol $x$ to mean a real number in this system, we regard it as the number $x + 0i,$ that is, the number whose real part is $x$ and imaginary part is $0.$
So, when we refer to a complex zero of a function, we mean a number $z$ such that $f(z) = 0$. While this doesn't seem to fit, we need to remember that the symbol $0$ is really the complex number that serves as the additive identity in $\mathbb{C}.$ So, we mean $0 = 0 + 0i$ as the complex number whose real and imaginary parts are both zero.
A: I don't quite like the other answer. We do not merely "regard" a real $x$ as $x+0i$. Neither should we think of "$0$" as merely a symbol representing "the complex number that serves as the additive identity in $ℂ$".
In actual fact, when we construct $ℂ$ we want $ℝ ⊆ ℂ$ literally. The claim is that there exists a field $ℂ$ that directly contains $ℝ$ and has some $i∈ℂ$ such that $i^2+1 = 0$. Here, both the $0$ and $1$ are exactly the same zero and one in $ℝ$! That is the whole point of complex numbers! It is non-trivial, but such a field exists and when we write "$ℂ$" we literally mean such a field!
So $0 = 0+0i$, and there is no "interpretation" or "regarding" at all. We define a complex zero of a function $f$ on $ℂ$ to be some $z∈ℂ$ such that $f(z) = 0$. I want to emphasize again that this $0$ is exactly the same zero as the natural number $0$, or the real number $0$.
The only reason we call it "complex zero" is to distinguish it from "real zero", which refers to some $x∈ℝ$ such that $f(x) = 0$. Still the same $0$! The only difference is that we are looking for real $x$.
