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zero of a function
Suppose $X$ is a set and $f$ a complexvalued function $f\colon X\to\mathbb{C}$. Then a zero of $f$ is an element $x\in X$ such that $f(x)=0$. It is also said that $f$ vanishes at $x$.
The zero set of $f$ is the set
$Z(f):=\{x\in X\mid f(x)=0\}.$ 
Remark. When $X$ is a “simple” space, such as $\mathbb{R}$ or $\mathbb{C}$ a zero is also called a root. However, in pure mathematics and especially if $Z(f)$ is infinite, it seems to be customary to talk of zeroes and the zero set instead of roots.
Examples

For any $z\in\mathbb{C}$, define $\hat{z}:X\to\mathbb{C}$ by $\hat{z}(x)=z$. Then $Z(\hat{0})=X$ and $Z(\hat{z})=\varnothing$ if $z\neq 0$.

Suppose $p$ is a polynomial $p\colon\mathbb{C}\to\mathbb{C}$ of degree $n\geq 1$. Then $p$ has at most $n$ zeroes. That is, $Z(p)\leq n$.

If $f$ and $g$ are functions $f\colon X\to\mathbb{C}$ and $g\colon X\to\mathbb{C}$, then
$\displaystyle Z(fg)$ $\displaystyle=$ $\displaystyle Z(f)\cup Z(g),$ $\displaystyle Z(fg)$ $\displaystyle\supseteq$ $\displaystyle Z(f),$ where $fg$ is the function $x\mapsto f(x)g(x)$.

For any $f\colon X\to\mathbb{R}$, then
$Z(f)=Z(f)=Z(f^{n}),$ where $f^{n}$ is the defined $f^{n}(x)=(f(x))^{n}$.

If $f$ and $g$ are both realvalued functions, then
$Z(f)\cap Z(g)=Z(f^{2}+g^{2})=Z(f+g).$ 
If $X$ is a topological space and $f:X\to\mathbb{C}$ is a function, then the support of $f$ is given by:
$\operatorname{supp}f=\overline{Z(f)^{\complement}}$ Further, if $f$ is continuous, then $Z(f)$ is closed in $X$ (assuming that $\mathbb{C}$ is given the usual topology of the complex plane where $\{0\}$ is a closed set).
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