MHB Mapping/determining domain/range

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find the range for
$g(z) = z^2$ for $z$ in the first quadrant, ie $Re z > 0 $ and $Im z > 0$

Why is the answer $Im w > 0$.

Similarly, how do i go about finding the range for:

$p(z) = -2z^3$ for $z$ in the quarter disk $|z|<1$, $0<Arg z<\frac{\pi}{2}$

I am confused as to how to determine the answer, what is the methodical approach to tackle this problem?
thanks.
 
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nacho said:
find the range for
$g(z) = z^2$ for $z$ in the first quadrant, ie $Re z > 0 $ and $Im z > 0$

Why is the answer $Im w > 0$.

Setting $\displaystyle z = \rho\ e^{i\ \theta}$ is...

$\displaystyle g(\rho, \theta) = \rho^{2}\ e^{2\ i\ \theta} = \rho^{2}\ (\cos 2\ \theta + i\ \sin 2\ \theta ) (1)$

... and because is $\displaystyle 0 \le \theta \le \frac{\pi}{2}$ is...

$\displaystyle \sin 2 \theta \ge 0 \implies \text{Im}\ (z^{2}) \ge 0$...

Kind regards

$\chi$ $\sigma$
 
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

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