Why treat complex scalar field and its complex conjugate as two different fields? I am new to QFT, so I may have some of the terminology incorrect.
Many QFT books provide an example of deriving equations of motion for various free theories. One example is for a complex scalar field:
$$\mathcal{L}_\text{compl scaclar}=(\partial_\mu\phi^*)(\partial^\mu\phi)-m^2\phi^*\phi.$$
The usual "trick" to obtaining the equations of motion is to treat $\phi$ and $\phi^*$ as separate fields. Even after this trick, authors choose to treat them as separate fields in their terminology. This is done sometimes before imposing second quantization on the commutation relations, so that $\phi$ is not (yet) a field of operators. (In particular, I am following the formulation of QFT in this book by Robert D. Klauber, "Student Friendly Quantum Field Theory".)
What is the motivation for this method of treating the two fields as separate? I intuitively want to treat $\phi^*$ as simply the complex conjugate of $\phi,$ not as a separate field, and work exclusively with $\phi$.
Is it simply a shortcut to obtaining the equations of motion
$$(\square +m^2)\phi=0\\
(\square + m^2)\phi^*=0~?$$
I also understand that one could write $\phi=\phi_1+i\phi_2$ where the two subscripted fields are real, as is done here; perhaps this addresses my question in a way that I don't understand.
 A: TL;DR: Yes, it is just a short-cut. The main point is that the complexified map
$$\tag{A} \begin{pmatrix} \phi \\ \phi^{*} \end{pmatrix}
 ~=~ \begin{pmatrix} 1 & i\\ 1 &-i \end{pmatrix}
\begin{pmatrix} \phi_1 \\ \phi_2 \end{pmatrix} $$
is a bijective map :$\mathbb{C}^2 \to \mathbb{C}^2 $. 
Notation in this answer: In this answer, let $\phi,\phi^{*}\in \mathbb{C}$ denote two independent complex fields. Let $\overline{\phi}$ denote the complex conjugate of $\phi$.
I) Let us start with the beginning. Imagine that we consider a field theory of a complex scalar field $\phi$. We are given a Lagrangian density 
$$\tag{B} {\cal L}~=~{\cal L}(\phi,\overline{\phi},\partial\phi, \partial\overline{\phi})$$ 
that is a polynomial in $\phi$, $\overline{\phi}$, and spacetime derivatives thereof. We can always decompose a complex field in real and imaginary parts
$$\tag{C} \phi~\equiv~\phi_1+ i \phi_2 ,$$
where $\phi_1,\phi_2 \in \mathbb{R}$. Hence we can rewrite the Lagrangian density (B) as a theory of two real fields
$$\tag{D}{\cal L}~=~{\cal L}(\phi_1,\phi_2,\partial\phi_1, \partial\phi_2).$$
II) We can continue in at least three ways:


*

*Vary the action wrt. the two independent real variables $\phi_1,\phi_2 \in \mathbb{R}$.

*Originally $\phi_1,\phi_2 \in \mathbb{R}$ are of course two  real fields. But we can complexify them, vary the action wrt. the two independent complex variables $\phi_1,\phi_2 \in \mathbb{C}$, if we at the end of the calculation impose the two real conditions
$$\tag{E} {\rm Im}(\phi_1)~=~0~=~{\rm Im}(\phi_2). $$

*Or equivalently, we can replace the complex conjugate field $\overline{\phi}\to \phi^{*}$ in the Lagrangian density (B) with an independent new complex variable $\phi^{*}$, i.e. treat $\phi$ and $\phi^{*}$ as two independent complex variables, vary the action wrt. the two independent complex variables $\phi,\phi^{*} \in \mathbb{C}$, if we at the end of the calculation impose the complex condition
$$\tag{F} \phi^{*} ~=~ \overline{\phi}. $$
III) The Euler-Lagrange equations that we derive via the two methods (1) and (2) will obviously be exactly the same. The Euler-Lagrange equations that we derive via the two methods (2) and (3) will be just linear combinations of each other with coefficients given by the constant matrix from eq. (A).
IV) We mention for completeness that the complexified theory [i.e. the theory we would get if we do not impose condition (E), or equivalently, condition (F)] is typically not unitary, and therefore ill-defined as a QFT. Recall for starter that we usually demand that the Lagrangian density is real. 
References:


*

*Sidney Coleman, QFT notes; p. 56-57.

A: I would like to make a comment, which may clarify and simplify the things a little bit.
In complex analysis [see e.g. ``Introduction to Complex Analysis" by B.V. Shabat] by definition  derivatives over the complex variables $z$ and $\bar z$ are given by: 
$$
\mbox{def:} \quad \partial_z  \equiv \frac{1}{2}
\left(\partial_{\rm a} - i \partial_{b}\right) \quad  
\partial_{\bar z}  \equiv \frac{1}{2}
\left(\partial_{\rm a} + i \partial_{b}\right),
$$
where $a$ and $b$ stand for real and imaginary parts of $z$ correspondingly.
The equalities 
$$\partial_{ z} \bar z = 0 \quad \mbox{and} \quad \partial_{\bar z} z = 0 $$ 
imply, that the variations over z and $\bar z$ are independent, while    the variables $z$ and $\bar z$ (being mutually complex conjugated) are  not independent. There is no any doubling of degrees of freedom, but
one can vary over the field and its conjugated considering them as independent.
A: Of course @QMechanic's answer is correct.
i would like to show a very simple reason why this is so (and also point to possible generalisations)
First of all, any complex number $z=a+bi$, is 2-dimensional and each part (the real part $a$ or the imaginary part $bi$) can be completely independent of each other. As a result a complex number can represent in condensed form 2 numbers. Moreover this also means that a complex number to be completely determined each of the dimensions needs to be determined as well.
On the other hand, from every complex number $z=a+bi$ (along with its complex conjugate $\bar{z}=a-bi$), one can calculate 2 real numbers ($a$, $b$) as:
$$a = (z + \bar{z})/2$$
$$b = (z - \bar{z})/{2i}$$
Since $a$ and $b$ can be completely independent of each other, so can $z$ and $\bar{z}$.
There is a complete symmetry of representation (if such a term can be used).
This means that in QFT (for example), instead of doing variations on the $a$, $b$ real fields, one can equivalently (by the same token) do variations on the $z$, $\bar{z}$ complex fields and so on.
UPDATE:
To get into the abstract mathematics a little more.
Complex conjugation is (the natural) automorphism of the field of complex numbers. Furthermore the complex conjugate of a complex number $z$ can not be derived from any analytic function of $z$ (roughly meaning rational functions of $z$ and power series). This further makes the complex conjugate $\bar{z}$ natural candidate for treating as separate field.
Quiz:  How many components are needed to compute the velocity $v=dx/dt$ of an object having position $x$, and can these be considered independent? Or in other words knowing position $x$ (at a given time $t$), can we also know velocity $v$ (at the same given time)??
