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I want to show that the mapping given by $$w = f(z) = - \frac{1}{2} \left( z + \frac{1}{z} \right)$$ is a bijective mapping from the upper half disc to the upper half plane.

One-to-one case is straightforward. For onto, for $w \in \mathbb{C}$ such that $\text{Im}(w) > 0$, how to show that the value $$z = \sqrt{w^2 - 1} - w $$ is in the upper-half of the unit disk, i.e., $|z| < 1$ with $\text{Im} (z) > 0$.

I tried substituting $w = u + iv $ with $v > 0$ but wasn't successful. Any help is much appreciated.

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    $\begingroup$ show that the boundaries match as that is fairly straightforward and this gives onto $\endgroup$
    – Conrad
    Commented Dec 29, 2021 at 23:02
  • $\begingroup$ @Conrad, true but the upper half-plane has the real axis as the boundary while the boundary of the upper half-disc consists of a line segment and an arc which makes it difficult to show the mapping. $\endgroup$
    – Eureka
    Commented Jan 2, 2022 at 5:52
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    $\begingroup$ Not really as there are $3$ special points on the real axis ($\pm 1, \infty$) which will guide you $\endgroup$
    – Conrad
    Commented Jan 2, 2022 at 14:33
  • $\begingroup$ @Conrad, yes 3 points determine a conformal map. Could you pls provide a proof. Thanks ! $\endgroup$
    – Eureka
    Commented Jan 3, 2022 at 1:12
  • $\begingroup$ show that $1 \to -1, -1 \to 1, \pm \infty \to 0, (-1,1) \to |z|=1, \Im z >0, (1, \infty) \to (-1,0), (-\infty, -1) \to (0,1)$ which are fairly straightforward all (eg $-1<x<1, \sqrt {x^2-1}=i\sqrt{1-x^2}$ so $|-x+ \sqrt {x^2-1}|=1, \Im ()>0$ and conversely any such $w$ is uniquely $-x+ \sqrt {x^2-1}$ for $-1<x<1$; the others are really simple noting that for $x<-1, \sqrt {x^2-1}<0$ so $0< \sqrt {x^2-1}-x<1$ while for $x>1, \sqrt {x^2-1}>0$ so $-1< \sqrt {x^2-1}-x<0$ $\endgroup$
    – Conrad
    Commented Jan 3, 2022 at 2:20

1 Answer 1

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For onto, for $w \in \mathbb{C}$ such that $\text{Im}(w) > 0$, how to show that $\;$...$\;$ $|z| < 1$ with $\text{Im} (z) > 0$.

The equation $\,z^2 + 2 w\,z + 1 = 0\,$ is a quadratic in $z$, which has two complex roots with product equaling $\,1\,$ by Vieta's relations, so one of them must be inside the unit circle. The question then reduces to showing that $\,z\,$ is in the upper-half of the unit disk when $w$ is in the upper half plane.

Using that $\displaystyle\,\frac{1}{z}=\frac{\bar z}{|z|^2}\,$ and $\,a - \bar a = 2i\,\text{Im}(a)\,$:

$$ -2 \cdot 2i\, \text{Im}(w)=z + \frac{\bar z}{|z|^2}-\left(\bar z + \frac{z}{|z|^2}\right)=\left(z-\bar z\right)\left(1 - \frac{1}{|z|^2}\right) = 2i\,\text{Im}(z)\,\left(1-\frac{1}{|z|^2}\right) \\ \implies\;\;\;\; \underbrace{2\, \text{Im}(w)}_{\text{w in the upper} \\ \text{half-plane}} = -\,\underbrace{\underbrace{\text{Im}(z)}_{\gt 0} \bigg/ \underbrace{\left(1-\frac{1}{|z|^2}\right)}_{\lt\,0}}_{\text{z in the upper-half} \\ \text{ of the unit disk}} \;\gt\; 0 $$

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  • $\begingroup$ thank you ! Can you please tell me how does the first equality follows ? Also, in the conclusion you've taken $|z| < 1$ which is one condition that we're supposed to show. $\endgroup$
    – Eureka
    Commented Jan 16, 2022 at 9:34
  • $\begingroup$ @Eureka Thanks, there was a typo, edited and fixed now. Not sure which first equality. $\,z^2 + 2 w\,z + 1 = 0\,$ follows straight from the given equation, and so does $\,-2 \cdot 2i\, \text{Im}(w)=...$ $\endgroup$
    – dxiv
    Commented Jan 16, 2022 at 10:00

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