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Let $L_1$ be the $x$-axis, let $L_2$ be the $y$-axis and let $L_3$ be the vertical line $x=1$. For each $k \in \mathbb{Z}$ let $C_k$ denote the circle of radius $r=\frac{1}{2}$ with centre $z=\frac{1}{2}+ki$. Let $f(z)=\frac{2z}{z+1}$.

How would I go about finding the images $f(L_1)$, $f(L_2)$, $f(L_3)$ and $f(C_0)$?

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    Just checking: Have you tried parametrizing each curve, substituting the parametrization into $f$, and finding the real and imaginary parts of the resulting value? (It will help to perform polynomial division on $f$ first.) – 2017-01-07
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    Linear fractional transformations take circles and lines to circles and lines (a circle can go to a line and vice versa). Circles are bounded and lines are unbounded. A circle is compact so it can only go to a line if it has some point $z_0$ such that $f(z_0)=\infty$. – 2017-01-07
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    @AndrewD.Hwang How would I parametrize each curve? – 2017-01-07
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    The line $x = c$: by $\gamma(t) = (c, t)$. The circle of radius $r$ and center $z_{0}$: By $\gamma(t) = z_{0} + re^{it}$. – 2017-01-07
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    @AndrewD.Hwang I'm afraid I'm still stuck on this. Please can you show me how to parametrize $L_1$ and I will try and figure out how to do the rest? – 2017-01-08
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    The $x$-axis can be parametrized by $\gamma(t) = (t, 0)$ (or, as a complex number, by $\gamma(t) = t$), with $t$ real. – 2017-01-08
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    So $f(L_1)= \frac{2t}{t+1}$? As this is always real, and $f(-1) = \infty$ it is a line i.e. the $x$-axis? It maps $L_1$ to itself? – 2017-01-08
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    Sorry, forgot to tag you @AndrewD.Hwang. Thanks for all your help by the way. – 2017-01-08
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    You're very welcome, and yes, that's exactly right. :) – 2017-01-08
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    @AndrewD.Hwang The line $x=0$ can be parametrised by $\gamma (t) = (0,t)$. $f(L_2) = \frac{2it}{it+1}$. I'm a bit confused by this as $f(1),f(-1)$ and $f(-i)$ all lie on the unit circle, but $f(i) = \infty$ suggesting this is a line? – 2017-01-08
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    Note that $\pm 1$ don't lie on the $y$-axis. That is, once you've set $z = it$ with $t$ real, $z = \pm1$ do not arise as inputs of $f$. (Separately, it probably helps to write $f(z) = 2 - \frac{2}{z + 1}$.) – 2017-01-08
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    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/51413/discussion-between-serveoverice-and-andrew-d-hwang). – 2017-01-08

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