Consider the equation $z^2 + 1$. As you probably know, $i$ is a root of this equation. In fact, this is the most commonly used definition of $i$: a number such that $i^2=-1$.
But $-i$ is also a root of $z^2+1$, so why not choose this one? Ok. Let us write $i'=-i$. Since $i$ and $i'$ share the same definition, "everything" that is satisfied by one of them has to be satisfied by the other one. For example, since $(1+i)^4 = -4 $, we must have $(1+i')^4 = -4$; this is true!
When I say "everything", I actually mean every polynomial equation. The general principle is the following: if $i$ is a root of some polynomial $P(z)$ with real coefficients, then so $i'=-i$. In the previous example, the polynomial would be $P(z) = (1+z)^4 + 4$.
This principle is more powerful than it looks at first sight. For example, if $a+ib$ (with $a,b \in \Bbb R$) is a root of some polynomial $Q(z)$ with real coefficients, then $i$ is a root of $P(z) = Q(a+zb)$. We deduce that $-i$ is a root of $P(z)$, that is $a-ib$ is a root of $Q(z)$. In the same way, you can see that
$$
(a_1+ib_1) + (a_2+ib_2)=(a_3+ib_3) \iff (a_1-ib_1) + (a_2-ib_2)=(a_3-ib_3)
$$
with $P(z) = (a_1+zb_1) + (a_2+zb_2)-(a_3+zb_3)$, and
$$
(a_1+ib_1)(a_2+ib_2)=(a_3+ib_3) \iff (a_1-ib_1)(a_2-ib_2)=(a_3-ib_3)
$$
(which polynomial would you take?)
The notion of conjugate is just a way to formalize all these "symmetries" between $i$ and $-i$. For instance, the two last equivalences write
$$
\overline{z_1}+ \overline{z_2} = \overline{z_1+ z_2}
,\qquad
\overline{z_1}\times \overline{z_2} = \overline{z_1\times z_2}
$$
Remark: the same kind of things could be done with $\sqrt{2}$ and $-\sqrt{2}$ and polynomial equations with rational coefficients. The general idea behind all of this is is that of Galois Theory.