Author Topic: 3.3 Q7, b,c,d  (Read 48 times)

Ende Jin

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3.3 Q7, b,c,d
« on: December 05, 2018, 08:28:32 PM »
I have tried (b) and had some clue about it. Basically I choose the intersection of the original to map to the intersection in the codomain. That means choosing $\{0, \infty\} \mapsto \{0, 1\}$. After that, I can get only a form like $z \mapsto \frac{az}{az+d}$. And then I give it $x \in \mathbb{R}$ and $iy, y \in \mathbb{R}$ to ensure the two condition is satisfied. With that, I can get $ad \in \mathbb{R},\bar{a}d \in \mathbb{R}$. And I arbitrarily chose  a solution and it works.

My question is if there is a general way to do this kind of question? Because in the above, there are too many guessing.

Ende Jin

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Re: 3.3 Q7, b,c,d
« Reply #1 on: December 05, 2018, 08:44:34 PM »
Also, is it true that if a mobius transformation maps a circle/a line to a circle/line, then the points in one area in the domain (because there are totally two areas) that is on one side of the circle/line will always maps onto the area in the same side of the circle/line in the codomain?