geometrical multiplication of complex numbers It is taught that $\mid z \mid ^{2} =x^{2}+y^{2}$ and $\bar{z}z=x^{2}+y^{2}$. Algebrically its fine to understand it. But what is the geometrical meaning of it? I tried multiplying couple of numbers but didn't find some good reason.
 A: Consider the polar form of a complex number.
$$z = |z| e^{i \theta}$$
When you multiply two complex numbers, the magnitudes multiply and the angles add.  Let $w = |w| e^{i \phi}$, so that
$$zw = |z||w| e^{i (\theta + \phi)}$$
Multiplying a complex number by another dilates and rotates according to these values.
A: 
when you multiply $\bar{z}$ by $z$ you are rotating $\bar{z}$ for a angle $\theta$ where $-\theta$ is the angle of $\bar{z}$(with respct to axis x). So $\bar{z}$ is "going" to axis x(with the same norm).
A: Who knows if you will find this valuable, but this is a method I really like.
Identify each complex number $z = a + bi$ with the matrix $\left(\begin{array}{cc} a & -b \\ b & a \end{array}\right)$.  If you multiply matrices for two complex numbers $z_1$ and $z_2$, you obtain the matrix of the product $z_1$ and $z_2$.
You can deduce that for any complex number $z \neq 0$ has a representation for $z/|z|$ which is a rotation matrix.
I'll leave it to you (or another answer) to spell it out exactly, but you can think of complex numbers as having two geometric parts - their magnitude and their rotation component.
A: It's easier to visualize the action in polar coordinates, (particularly exponential form using Euler's formula $z= r e^{i \phi}$.)
Standard ($x y$) and conjugate ($x \bar{y}$) multiplication both act in the same way with respect to the modulus: they multiply the moduli. 
However, whereas standard multiplication adds the phases, conjugate multiplication subtracts the phases. 
