I am studying Garnett's Bounded Analytic Functions. At page 3, it is stated that if $\tau(z)=\frac{z-z_0}{1-z\overline{z_0}}$, then the straight line though $0$ and $z_0$ is invariant under $\tau$. I could not verify this statement. Please help.
Straight line invariant under a Mobius transformation
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complex-analysis
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0I could be wrong, but I don't think it's invariant for all $\theta$. $e^{i\theta}$ represents rotation, and a straight line cannot be invariant under rotations unless $\theta=0$ or $\theta=\pi$ – 2017-02-02
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0As @user160738 said, it's wrong in general. The line through $0$ and $z_0$ (where we suppose $z_0 \neq 0$) is fixed by $\tau$ if and only if $\theta$ is an integer multiple of $\pi$. – 2017-02-02
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0Sorry, I have mistaken. The $\theta$ is zero in the above statement. – 2017-02-02
1 Answers
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For fix $\theta$, let $z=\lambda z_0$ is the line oass through $0$ and $z_0$ so $$\tau(\lambda z_0)=e^{i\theta} \frac{\lambda z_0-z_0}{1-\lambda z_0\overline{z_0}}=z_0e^{i\theta} \frac{\lambda -1}{1-\lambda z_0\overline{z_0}}=z_0\lambda'$$ then the straight line is invariant under $\tau$.
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0Thank you. I got the point. I was having trouble with the $1-\lambda |z_0|^2$ part. I was confused. Thanks again. – 2017-02-02