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I'm trying to solve the following complex number problem:

Let

$$S=\{z\in\mathbb{C}:|z|=1\}$$ $$D=\{z\in\mathbb{C}:|z|\leq1\}$$ $$f: \mathbb{C}-\{3\}\rightarrow\mathbb{C}$$ $$f(z)=\frac{3z-1}{3-z}$$

Prove that $f(S)=S$, $f(D)=D$.

So far the only thing I managed to prove is that $f(S) \subset S$: $$|f(a+bi)|=\frac{|3a+3bi-1|}{|3-a-bi|}=\frac{\sqrt{9a^2-6a+1+9b^2}}{\sqrt{a^2-6a+9+b^2}}=\frac{\sqrt{9-6a+1}}{\sqrt{1-6a+9}}=1$$

But it seems I need to do something smarter in order to prove remaining inclusions. I feel like there is some known standard way to do such problems, I just don't know it. I would be helpful for any tips. I've also tried using trigonometric form, but to no avail.

Rebellos
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qiubit
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  • this might be helpful https://math.stackexchange.com/q/102659 – mate89 Nov 30 '17 at 00:22
  • This is a Möbius transformation, completely characterised by its action on three distinct points. So pick three points on the unit circle and see how it acts on them. Then pick one point in the interior and see where it goes. –  Nov 30 '17 at 03:08

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