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In order to show that the image of the extended real line under any Möbius transformation is either a circle or a straight line I consider a Möbius transformation $T$ and have managed to show that $w=T(z)$ is either lying on a circle in the Riemann sphere or on a straight line in the Riemann sphere if $z=\bar z$ i.e. if $z \in \mathbb R_{\infty}$. But it doesn't mean that the image of $\mathbb R_{\infty}$ under $T$ is a circle or a straight line i.e. $T$ maps $\mathbb R_{\infty}$ onto a circle or a straight line. It may so happen that the image of $T$ is a portion of a circle or a portion of a straight line in the Riemann sphere since $T$ is continuous in the Riemann sphere.

But then how can I reach towards the desired conclusion from here? Please help me.

Thank you in advance.

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Suppose the image of $\mathbb{R}_\infty$ is only part of a line or circle. Mobius transforms are also bijections, so the points of the circle or line that are not in the image of $\mathbb{R}_\infty$ must come from somewhere. So what are their preimages?

A very "use the one theorem" hint: The residue formula applies to $\int_{\mathbb{R}_\infty} \frac{\mathrm{d}z}{z-\mathrm{i}}$. It must also apply to the image of $\mathbb{R}_\infty$ and the image of the pole at $\mathrm{i}$, so what can you conclude about the image of the line?

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  • $\begingroup$ I dont understand the thing "The residue formula..." $\endgroup$ – Arnab Chattopadhyay. Nov 13 '17 at 6:28

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