Gemetric representation of complex numbers Find the geometric representation of;
|z-2| - |z+2| < 2
things i know;
| z - 2 | is the distance between a point z and the point (2,0) in the complex plane. It suppose to represent a hyperbola, BUT i have no idea on how to get there. i know the standard form of a hyperbola, but just, no idea on how to manipulate this expression into the standard form. 
 A: If you want to manipulate it algebraically (without considering the geometric definition of the hyperbola), then let $$z=x+iy$$ and note that $$|z|^2=z\overline{z}=x^2+y^2.$$
Edit
First, as someone mentioned in the comments, I am not sure if you want the equation of the hyperbola or the set of points that satisfy your relation. This is how you can algebraically obtain the equation of the hyperbola:
Consider $|z-2|-|z+2|= 2$
Then, letting $z=x+iy$ and using the definition of $|z|$ above, we have
\begin{align*}
|(x-2)+iy|-|(x+2)+iy|&=2 \\
\sqrt{(x-2)^2+y^2}-\sqrt{(x+2)^2+y^2}&=2 \\
(x-2)^2+y^2&=\left(2+\sqrt{(x+2)^2+y^2}\right)^2 \\
&=4+4\sqrt{(x+2)^2+y^2}+(x+2)^2+y^2.
\end{align*}
So
\begin{align*}
\left((x-2)^2-(x+2)^2-4\right)^2&=\left(4\sqrt{(x+2)^2+y^2}\right)^2 \\
\left(-8x-4 \right)^2&=16\left((x+2)^2+y^2\right)\\
16(4x^2+4x+1)&=16\left((x+2)^2+y^2\right) \\
4x^2+4x+1-(x+2)^2-y^2&=0 \\
3x^2-3-y^2&=0 \\
\end{align*}
So the equation of the hyperbola is 
$$x^2-\left(\frac{y}{\sqrt{3}}\right)^2=1$$
The geometric way is perhaps simpler:
The equation of the hyperbola is given as
$$|z-z_1|-|z-z_2|=2a,$$ where $z_1,z_2$ are the foci (remember $z=x+iy$ so we can assign the coordinates $(x,y)$ to $z$), and $a$ is half of the major axis. For our example we have $z_1=(2,0)$ and $z_2=(-2,0)$ and the focal length is $2$. Since the focal length is given by $\sqrt{a^2+b^2},$ where b is the minor axis, we have 
$$2^2=1+b^2,$$ so $b= \sqrt{3}$. Putting all of this together gives the same equation as above.
Note
Technically we want $$||z-z_1|-|z-z_2||=2a,$$ in order to ensure that the distance does not become negative  (if we go to the other branch).
Edit
To solve your inequality, we do
\begin{align*}
x^2-\left(\frac{y}{\sqrt{3}}\right)^2&<1 \\
y^2&>3(x^2-1),
\end{align*}
So
$$y>\sqrt{3(x^2-1)}$$ or
$$y<-\sqrt{3(x^2-1)}$$
