Some clarification of complex projection of $v_1$ onto $v_2$? Suppose $v_1$ and $v_2$ are $2$-D real vectors, then we can easily define projection of $v_2$ along $v_1$ as
$$
\textbf{Proj}_{v_1}(v_2) = \frac{\langle v_2,v_1\rangle}{\|v_1\|} \cdot \frac{v_1}{\|v_1\|} =  \frac{\langle v_2,v_1\rangle}{\|v_1\|^2}v_1
$$
First question:
We can write either $\langle v_2,v_1\rangle$ or $\langle v_1,v_2\rangle$ when $v_1$ and $v_2$ are $2$-D real vectors. However, when $v_1$ and $v_2$ are $2$-D complex vectors, it matters if we write one or the other. Which one is correct? Please rigorously justify that.
Second question:
We define projection of a vector $v_2$ onto a subspace $W$ in general not along the direction of the other vector ($v_1$) and write $\textbf{Proj}_{W}(v_2)$. Can we say $W=\text{span}\{v_1\}$ in this case? if this is the case how one would define complex inner product for two complex vectors to define the above projection?
 A: Note that if we speak about complex vector the inner product if often called the hermitian product (you can find key differences here) which acts by conjugaction in the sense that $\langle v_{1},v_{2}\rangle = \overline{\langle v_{2},v_{1}\rangle}$ (where the bar denotes the conjugated) and it depends wether you want to project $v_{2}$ 'along' $v_{1}$ or vice-versa.
In order to clarify things the real case might help. 
What does the term 'along' refers to ? In depends on the definition you can easily pass from one to other once understood the concept behind. 
Generally it refers 'along' the orthogonal space spanned of $v$. We know that if the vector $v$ is such that $\langle v,v\rangle \ne 0$ (which means that $v$ is not an isotropic vector) we have a 'privileged' decomposition of vectorial space as $V = span(v) \oplus span(v)^{\perp}$. 
(In general if $W$ is a subspace of the vectorial space the decomposition $V = W \bigoplus W^{\perp}$ holds as long as the radical of the scalar product restricted to the subspace $W$ is not degenerate). 
The question we want to ask is how a decomposition such as $V = span(v) \oplus span(v)^{\perp}$ can affect the form of a certain $w \in V$?
Since $w \in W$,if the decomposition holds, it must exists $\lambda \in \mathbb{R}, z \in span(v)^{\perp}$ such that $w = \lambda v + z$. 
If we want to find $\lambda$ we can proceed as follows, since we know $z = w - \lambda v$ and $z \in span(v)^{\perp}$ it must holds that $0 = \langle v,z \rangle = \langle w-\lambda v,v \rangle = \langle w,v \rangle - \lambda \langle v,v \rangle $ obtaining $\lambda = \frac{\langle w,v \rangle}{\langle v,v \rangle}$
(Notice that it is well define since $v$ is not isotropic)
We discover that $\forall w \in V, w = \frac{\langle w,v \rangle}{\langle v,v \rangle} v + z$
$\lambda = \frac{\langle w,v \rangle}{\langle v,v \rangle}$ is often called the 'Fourier coefficient', and note also that $\lambda$ give us the coordinates in this case in $\mathbb{R}^{n}$ of the vectors in $V$, given an 'orthogonal' decomposition.
To answer your question this is what there is behind, if you want an heuristic interpretation what we are doing is projecting our vector on the 'direction' given by the orthogonal given by the decomposition of $V$.
Some reference of the theorem can be found here
