How Represent Waves via Complex Numbers? i try to finished my thesis, (Just have a problem with the wave mechanics)
this is wave function: 
$$\Psi(\vec x, t)=A\exp{i(\phi+\vec k.\vec x-\omega t)}$$
In mathematics, the symbol $i$ is conventionally used to represent the square-root of minus one: i.e., $i^2=-1$
In addition, a general complex number is written:
$$z=x+iy$$
$$z=r|\cos\theta + i\sin\theta|=r\exp{i\theta}$$
that which means:
$$\Psi(\vec x, t)=A|\cos(\phi+\vec k.\vec x-\omega t) + i\sin(\phi+\vec k.\vec x-\omega t)|$$
ok i want to know that:


*

*$$A\cos(\phi+\vec k.\vec x-\omega t)=?$$

*$$A\sin(\phi+\vec k.\vec x-\omega t)=?$$


in the case of complex numbers is:
1. $$r\cos\theta=x$$
2. $$r\sin\theta=y$$
and also i want to know that what is unit of $A$ ?
 A: Re the units of $A$: $A$ has dimensions of $L^{-3/2}$. The modulus of $\Psi^2(\vec{x},t)$ is a probability density so it has units of probability per unit volume, i.e. it's dimension is $L^{-3}$, so the dimensions of $\Psi$ and therefore $A$ must be $L^{-3/2}$. I don't think there is an SI unit corresponding to $\sqrt{P/m^3}$, but then it's hard to see when you would need such a unit.
Re the other question: the function:
$$\Psi(\vec x, t)=Ae^{i(\phi+\vec k\vec x-\omega t)}$$
is indeed a solution to the Schrodinger equation for a free particle, but you need to be careful about it's physical meaning. As mentioned above, the modulus of $\Psi^2$ gives you the probablity of finding the particle per unit volume, but $\Psi$ itself doesn't have a physical interpretation. You're quite correct that the equation can also be written:
$$\Psi(\vec x, t)=Acos(\phi+\vec k\vec x-\omega t) + Aisin(\phi+\vec k\vec x-\omega t)$$
but you can't attach any physical meaning to the two terms on the right.
P = probability
Re the comment: I suppose the SI unit would be the metre$^{3/2}$, but the value of the wavefunction has no physical significance so no such unit has ever been defined.
This is why the value has no physical significance: suppose $\phi$ is zero, then what is $\Psi(0, 0)$. The answer is of course $A$. But the phase $\phi$ has no absolute meaning, it's a phase relative to some reference point that can be freely selected. So suppose I use a reference point that is $\pi/2$ different to yours then I'd calculate $\Psi(0, 0)$ to be $iA$ not zero.
So you think $\Psi(0, 0) = A$ while I think $\Psi(0, 0) = iA$. Does this make any difference in the real world. Well the probability deniity is $\Psi^*\Psi$ so you calculate the probablility density to be $A^2$, and I calculate the probability density to be, erm, $A^2$ and we get the same answer even though we have different values for $\Psi$.
