MHB Mapping/determining domain/range

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The discussion focuses on determining the range of the function g(z) = z^2 for z in the first quadrant, concluding that the range is characterized by Im(w) > 0 due to the properties of the sine function in that quadrant. The method involves expressing z in polar coordinates, leading to the conclusion that the imaginary part of g(z) remains positive. Additionally, the range for p(z) = -2z^3 in the quarter disk is also explored, with participants seeking a systematic approach to find the range. The conversation emphasizes the importance of understanding the geometric implications of complex functions in specified domains. Overall, the thread highlights the need for clarity in mapping complex functions to their ranges.
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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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