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Find Möbius transformation that maps $x-$axis to $y=x$, maps $y-$axis to $y=-x$, maps the unit circle to itself.

I was thinking this means that $f$ maps $0$ to $0$, and maps 1 or -1 to $1+i$ or $1-i$, and maps $i$ or $-i$ to $-1+i$ or $1-i$. And I tried to plug in the formula $$f(z)=\frac{az+b}{cz+d}$$ But since there are so many combinations I am not sure I got the idea correct or not. Since I tried one combination and the result seems wrong...

Thanks for any help~

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Your idea is certainly doomed to failure, since $1$, $-1$, $i$, and $-i$ are all on the unit circle and so must map to points on the unit circle.

You could more carefully choose where you are mapping your points to make sure they are compatible with all the requirements. However, in this case there is a much simpler approach. Try just drawing a picture and thinking geometrically. What simple geometric operation would map the $x$-axis to the line $y=x$ and the $y$-axis to the line $y=-x$? The answer is hidden below.

A counterclockwise rotation of $\pi/4$ about the origin would map the lines as required, and also maps the unit circle to itself. As a Möbius transformation, this is $z\mapsto e^{i\pi/4}z$.

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  • $\begingroup$ Got it! Thanks! $\endgroup$ – Nan Dec 29 '17 at 20:49

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