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Given the LFT for a complex $z$, \begin{align*} \phi:z\mapsto \frac{2z+1}{z+2}. \end{align*} I'm asked about the image under $\phi$ of $C:=\left\{\left\lvert z+\frac25\right\rvert = \frac25\right\}$. I've parametrized this as $\gamma: \frac25 \exp(it) - \frac25$ and computed the image \begin{align*} w(t) = 2-\frac{15}{2\exp(it)+8}, \end{align*} but I'm none the wiser from this expression and I also don't see why this doesn't give me a new circle, although $\phi$ is conformal $ad-bc = 4 - 1 = 3 \neq 0$. Any hints?

1010011010
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2 Answers2

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Hint: Mobius transformations map generalized circles to generalized circles.

A generalized circle is either a 'regular' circle, or a straight line plus a point at infinity.

Pick three distinct points on the circle you are given, and find their image under your map. It should be evident if the image is a line or a circle.

For example, you can see that your map sends $0$ to $1/2$ and $−4/5$ to $3/2$ and $−2/5+2/5i$ to $4/17+15/34i$. These points definitely lie on a cricle

GaussTheBauss
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HINT...Write $w=\frac{2z+1}{z+2}$ and rearrange so you have $$z=\frac {1-2w}{w-2}$$

Then the locus becomes, after some simplification, $$|1-8w|=2|w-2|$$

Putting $w=u+iv$ will give the Cartesian equation of the circle.

David Quinn
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    I'm not sure what you mean – David Quinn Dec 26 '15 at 22:15
  • Ah sorry, my comment didn't make much sense. Your answer is very probably the easiest path to the solution; I figured that there was some kind of alternative path, but I again got stuck real soon. – 1010011010 Dec 27 '15 at 08:11
  • I have a small followup question on the matter: your solution provides a beautiful way to quickly obtain the (implicit) Cartesian expression for $w$ in terms of $u$ and $v$. My solution instead gives $w(t)$. Both expression contain the same geometric shape of the image, but my solution (as given in the OP) gives a different speed at which we go through the circle. Is there any way I can adjust my $t$ so that I can make the same substitution $\gamma$ as the one I gave in the OP? – 1010011010 Dec 27 '15 at 16:58
  • I'm not sure what you mean by different speed. You have obtained in $w(t)$ what can be expressed as the parametric form of the circle, whereby the real and imaginary parts of $w$ can be expressed in terms of $\cos t$ And $\sin t$ – David Quinn Dec 27 '15 at 17:08
  • Indeed, and although $\lvert w\rvert = \frac12$ (at least that's what I got from your hint), it should give some kind of expression in the form $w = \frac12 \cos \theta + \frac12 \sin \theta$, but the relation between $t$ and $\theta$ is unclear to me, and neither is it clear to me how to obtain the relation, which is what my question was getting at. – 1010011010 Dec 27 '15 at 22:09
  • Yes the connection seems obscure but intersting. I get $$\cos\theta=\frac{8+17\cos t}{17+8\cos t}$$ and $$\sin \theta=\frac{15\sin t}{17+8\cos t}$$ – David Quinn Dec 27 '15 at 22:48
  • There seems to be very little one can do to get a nice formula for $\theta(t)$, even with a priori knowledge of the shape of the circle. I think the (next) best way to do it is by noting that $\frac12$, $\frac12 i$, $-\frac12$, $-\frac 12i$ are all solutions of $w(t)$, and noting that the output can only be a circle, therefore the output is a circle of radius $\frac12$. – 1010011010 Dec 28 '15 at 08:11