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Let $f(z) = \dfrac{2z}{z+1}$ and $L_{3}$ = the line $x=1$.

I am required to find the image of $L_{3}$ under $f(z)$.

I have done this by choosing the three points $1, 1+i,\infty$ and seeing what they are mapped to.

I have ended up with $1, 2$ and $6/5+2i/5$. Is this another circle? If so what is the center and how would it be sketched?

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The image of a "circle" (counting lines as circles with infinite radius) under Moebius transformation is always a "circle" (again possibly a straight line) Since $∞$ is not part of your image (-1 would map there) this is definiatly a circle with finite radius. This could also be obtained by seeing, that the three points are not on a straight line.

Here are some possible ways to determine a circle from three points on it.

Edit: Another property of Moebius Transformation is the conservation of symmetry. So of you know, which points maps to infity, mirror it on your "circle" and find the "soon to be center". In this case, it is easy to see, that $-1$ maps to infinity. By mirroring this point on your line, you get $3$ as point, that will transform to the center of the circle. Plugging $z=3$ into your trnasformation gives you $1.5$.

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    I have used this method but have ended up with $x^2+y^2-x-y/5i=0$ and am struggling to see what the radius/center would be. I get a radius of 0 but I know this is not right, – 2017-01-13
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    WolframAlpha can do that too, it gives a "correct" result [link](http://www.wolframalpha.com/input/?i=circle+through+(1,0),(2,0)(6%2F5,+2%2F5)) – 2017-01-13
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Use analytic way. Let $z=1+iy$ and $w=u+iv$, so $$u+iv=\frac{2+2iy}{2+iy}$$ after simplifying you get $u^2-3u+v^2+2=0$ which specify the circle $$|z-\frac32|=\frac12$$ this shows center and radius of image.

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    I don't quite understand what you have done here, why did you use $z=\dfrac12+iy$? For my circle would I sub in $6/5+2i/5$ into $z$ for $|z-\frac32|=\frac12$? – 2017-01-13
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    I'm so sorry. I have mistaken, see update. – 2017-01-13