Calculating the codomain of a complex function How can I calcute the direct image of the fisrt quadrant under the complex function $f(z)=z^3$? or, in general of any complex function?
 A: Cartesian coordinates
$$
z^{3} = \left( x + i y\right)^{3} = 
\color{blue}{  \left(
x^3-3 x y^2
\right)}
+ i
\color{red}{\left(
3 x^2 y-y^3
\right)}
$$

Another perspective populating 25 points in mesh pattern and then applying the cubic transformation. The results are show here.

@Lubin asks for clarification of the above figure. The goal was to create a small grid of 25 points in the first quadrant. For example
$$
  z_{1} = 0 + 0i, \quad z_{2} = \frac{1}{2} + 0i, \quad z_{2} = 1 + 0i, \quad \dots, \quad z_{2} = 2 + 2i.
$$
Instead of using dots to represent the points, the index $k$ in $z_{k}$ was used. This is the image on the left. 
The image on the right is $z_{k}^{3}$, $k=1,25$. We can pick out a few patterns. Notice that the points where Im $z = 0$, points 1 through 5, all stay on this axis. The argument for these points is $0$, and tripling that angle leaves us at $0$. The points Re $z = 0$, the points 1, 6, 11, 16, and 21, all have argument of $\pi/2$ and map to $3\pi/2$. Where is point 25? The argument is $\pi/4$ which maps to $3\pi/4$.
Polar coordinates
As alluded by @ imranfat,
$$
  z^{3} 
  = \left( r e^{i\theta} \right)^{3} 
  = r^{3} e^{3i\theta}
$$
The figure below shows the mapping on a polar grid. Because of the cubic action, vectors outside the unit circle grow rapidly; those inside shrink rapidly. The angles triple. However the radius is $\left(2\sqrt{2}\right)^{2}$ which is outside of the display region.
If we define the domain Quadrant I as $r \in [0,\infty)$, $\theta \in [0, \pi / 2]$, the codomain is $r \in [0,\infty)$,  $\theta \in [0, 3\pi / 2]$.

