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What is the image of the set {$ { {z ∈ C : z = x + iy, x ≥ 0, y ≥ 0} } $} under the mapping $ z \to z^2$

my answer : $f(z) = z^2 =(x+iy)^2 = x^2-y^2 +2ixy$,

here I get $u=x^2-y^2$ and $v=2xy$. After that I am confused that how can I find the image of the set.

Please help me, as any Hints can be appreciated or if u have time u can tell me the solution. I would be more thankful.

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For fun:

Domain :

$D= ${$z| z=x+iy,$; $x\ge 0$, $ y\ge 0$}.

First quadrant in the complex plane.

Let $z= re^{i\theta}$, with

$r \in \mathbb{R^+}$; $0 \le \theta \le π/2.$

$f(z) := z^2= r^2e^{i2\theta}$,

$r \in \mathbb{R^+}$; $0\le 2\theta \le π.$

Image $f$: First and second quadrant in the complex plane.

Examples:

1) What is the image of the line :

$z= re^{i\theta}$ with $r \in \mathbb{R^+}$; $\theta = π/4$ ?

2) What is the image of a quarter circle:

$z=re^{i\theta}$ with $r= a$ (constant); $ 0 \le \theta \le π/2$ ?

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  • $\begingroup$ thanks a lots @Peter $\endgroup$ – user476275 Jan 24 '18 at 9:42
  • $\begingroup$ Michael.Welcome. $\endgroup$ – Peter Szilas Jan 24 '18 at 10:06
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Hint: The map $z \mapsto z^2$ is just the map $$re^{ i\theta}\mapsto r^2e^{ i2\theta}.$$

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  • $\begingroup$ namaskar Sahiba $\endgroup$ – user476275 Jan 24 '18 at 9:42
  • $\begingroup$ @Michael, Рада Вас видеть . $\endgroup$ – освящение Feb 3 '18 at 16:55
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Hint: At first I would suggest you sketch your set, call it $S$. That should be quite easy after the definition.

Well it is just the first quadrant.

Then consider points in the complex plain, represented by their polar-coordinate representation and figure out what the mapping $z \rightarrow z^2$ does to them.

Given $z$ this mapping scales z and doubles the angle from the positive "real line".

And now you should be able to imagine all points resulting of $S$ under this mapping.

Namely, any point in the upper halfplane can be reached through this mapping. The real line is obviously included.

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  • $\begingroup$ thanks Konstantin $\endgroup$ – user476275 Jan 24 '18 at 9:42

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