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.

The reason is also to acquire inverses and do division $$\frac{z}{q}=\frac{z\bar{q}}{q\bar{q}}=\frac{z\bar{q}}{|q|^2}$$ Where you get $(a+bi)(a-bi)=a^2-(bi)^2=a^2+b^2$