What is the motivation for complex conjugation?

One motivation, if you can call it that, is that $i^2=-1$ does not define $i$, because $-i$ also satisfies that equation.

So, there are two elements that could be $i$ and there is no algebraic reason for choosing one over the other. In other words, $\pm i$ are interchangeable, hence conjugation.

Technically, interchangeable means that there is an $\mathbb R$-automorphism of $\mathbb C$ interchanging $i$ and $-i$.

If $f(x)$ is a polynomial with real coefficients, and $z \in \mathbb C$ is a root of $f$, then $\overline{z}$ is also a root of $f$; in other words complex conjugation acts on the roots of $f$, and we can separate the roots of $f$ into orbits according to this action. An orbit is either a root with $z = \overline{z}$, i.e. a real root, or a pair $\{z, \overline{z}\}$ consisting of a non-real complex number and its complex conjugate. If $z_1, \dots, z_k$ are the real roots and $\{w_1, \overline{w_1}\}, \dots, \{w_r, \overline{w_r}\}$ are the pairs of complex-conjugate roots of $f$, it follows that $f$ factors as

$$f(x) = c \big((x-z_1)(x-z_2)\cdots(x-z_k)\big) \times \big((x^2-2\Re w_1 + |w_1|^2\big)\cdots(x^2-2\Re w_r + |w_r|^2\big))$$

All of the polynomials have real coefficients.

So we see that every polynomial with real coefficients factors as a product of linear factors and quadratic factors, all over the real numbers. All of this thanks to the existence of complex conjugation.