# Solve the following linear fractional transformation

Q. The bilinear transformation that maps the points $$z_1=2$$, $$z_2=i$$, $$z_3=-2$$ onto the points $$w_1=1$$, $$w_2=i$$, $$w_3=-1$$. Into what curve is the imaginary axis $$x=0$$ is transformed.

I am able to solve the first part of the question, that is $$W= \frac{3z+2i}{iz+6}.$$ Can you help me in solving the second part?

• It would improve your Question to define what (in the context of complex analysis?) you mean by a "bilinear transformation". It is possible you have a linear fractional transformation in mind? – hardmath Oct 25 '18 at 1:23
• @hardmath yes. I found at least one book, in English, that really does call them bilinear transformations. books.google.com/… – Will Jagy Oct 25 '18 at 1:32
• Hint: You have $W(z)$ defined as a "bilinear transformation" (aka linear fractional transformation, aka Moebius transformation). It is characteristic of such functions that they map lines or circles to lines or circles in the complex plane. The second part asks about the image of a line (the imaginary axis), so you should be able to identify "what curve" this is from three points on it. – hardmath Oct 25 '18 at 17:13

In fact, you don't have to find $$W$$. Suppose the imaginary axis is mapped to a circle $$\mathcal C$$ (not a generalized circle). $$\mathcal C$$ is orthogonal to the unit circle at $$i$$, therefore its center lies on the line $$\operatorname{Im} w = 1$$. On the other hand, $$-1$$ and $$1$$ are conjugate wrt $$\mathcal C$$, therefore its center lies on the real axis. Since this is impossible, we conclude that the imaginary axis is mapped to the straight line which is orthogonal to the unit circle at $$i$$.