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I am wondering how the image of the half plane $\{ z \in \mathbb{C} : \Re(z)>\frac{1}{4}\}$ looks like under the map $f: z\mapsto \sqrt{z}$. The branch of $\sqrt{\cdot }$ is taken such that $\arg(x)=0$ if $x\in (0,\infty)$. Is it possible to describe the image properly?

I know that the image of the plane $\{ z \in \mathbb{C} : \Re(z)>0\}$ is the sector which given in polar coordinates by $\{ (r,\theta) \in \mathbb{C} : \vert \theta \vert<\frac{\pi}{4}\}$. Thus the image of the set $\{ z \in \mathbb{C} : \Re(z)>\frac{1}{4}\}$ is contained in that sector.

Best wishes

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  • $\begingroup$ Ok. Can you try to tell me why that is the case? Or is there a picture of the image which shows how messy it is? $\endgroup$ – Sammyy Delbrin Apr 17 '17 at 5:10
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We will deal with the branching later

$$\begin{array}{rcl} \Re(z^2) &>& \dfrac14 \\ \Re(r^2e^{2i\theta}) &>& \dfrac14 \\ r^2\cos2\theta &>& \dfrac14 \\ r^2(\cos^2\theta-\sin^2\theta) &>& \dfrac14 \\ (x^2+y^2)\left(\dfrac{x^2}{x^2+y^2} - \dfrac{y^2}{x^2+y^2}\right) &>& \dfrac14 \\ x^2-y^2 &>& \dfrac14 \\ \end{array}$$

Thus, in polar coordinates, the image looks like $r^2 \cos 2\theta > \dfrac14$ and in Cartesian coordinates (Argand diagram), the image looks like $x^2 - y^2 > \dfrac14$.

Plotting gives:

The right hand side is the correct branch.

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  • $\begingroup$ Thank you for your answer. $\endgroup$ – Sammyy Delbrin Apr 17 '17 at 5:16

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