I realise this is a general question. I am self-teaching mathematics and I have observed that the zeros of real and complex functions are of much interest.
Question: Why are the zeros of real or complex so important?
I was going to say that the zeros completely determine a function because the function can be factorised into factors, with each factor corresponding to one of the zeros. For example, $f(x) = (x-1)(x-3)$ is completely determined by its zeros at $x=1$ and $x=3$.
However this doesn't work for functions like $g(x) = (x-1)(x-3) + 2$.
I appreciate your patience with a question that might be naive.
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$\begingroup$Yes, Weierstrass and Hadamard showed that "entire" functions (=holomorphic on the entire complex plane) do have product expansions in terms of their zeros, with some extra factors for convergence of that infinite product... and a leading exponential term.
$\endgroup$ 2 $\begingroup$Often the zeros carry special significance.
For example the location of non-trivial zeros of the Riemann-Zeta function has far reaching consequences in mathematics, especially for theories of prime numbers.
In ordinary linear differential equations, the zeros of these “functions of functions” are called the “homogenous solutions” and form a vector space of solutions (since every solution evaluates to zero you can take linear combinations and still end up with zero).
The study of the roots of polynomials led to Galois Theory which connected fields and groups and helped resolve longstanding issues of which polynomials are solvable and also issues about provability in geometry.
Finally, a trivial interest in zeros is that $f(x)=c \iff g(x):=f(x)-c=0$.
Just a smattering of examples to illustrate.
$\endgroup$ $\begingroup$A general remark: Solving any equation
$$A=B$$
is equivalent to solving
$$A-B =0$$
as soon as $A$ and $B$ are anything which at least vaguely resembles mathematical objects that can be subtracted from each other. In particular, if $A-B$ is something which you can express as a function, your task is now to
$$\text{ find the zeros of } A-B.$$
Hence, more often than one thinks, "solving an equation" immediately reduces to "finding zeros".
Now one could ask why solving equations is important, but I trust you don't ask that if you have ever worked in any STEM field. Or you and I trust people who claim that equations like$$\mathbf F + \mathbf{u} \frac{\mathrm{d} m}{\mathrm{d}t} = m {\mathrm{d} \mathbf v \over \mathrm{d}t}$$or$$\oint_C \mathbf{B} \cdot \mathrm{d}\boldsymbol{l} = \iint_S \left( \mu_0 \mathbf{J}+ \mu_0 \varepsilon_0 \frac{\partial \mathbf{E}}{\partial t}\right) \cdot \mathrm{d} \mathbf{S}$$or$$G_{\mu \nu} = R_{\mu \nu} - \frac{1}{2} R g_{\mu \nu}$$have importance, and it's useful to solve them occasionally, in order to hit a target or get a motor running or make a GPS useful or estimate how old the universe is.
$\endgroup$ 1 $\begingroup$Actually it does work perfectly well for $g(x)=(x-1)(x-3)+2$. This has zeros at $2\pm i$, and it is indeed true that$$g(x)=(x-2-i)(x-2+i).$$
The statement that the zeroes determine the function only holds for polynomials (and only up to multiplication by a constant). For polynomials this always works over the complex numbers, since any polynomial can be fully factorised over the complex numbers (this is the fundamental theorem of algebra), but not necessarily over the reals.
However, note that you may need to know the multiplicities of the roots, not just the roots themselves. $f_1(x)=x^3-x$ and $f_2(x)=x^3-x^2$ are not the same, even though they have the same roots $0$ and $1$, but really the roots of $f_1$ are $0,1,1$ and the roots of $f_2$ are $0,0,1$.
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