# Find Möbius transformation that send Re(z)=Im(z) to a circle and the real axis to itself

Problem 3.3.7d in Complex Variables, 2nd edition, by Stephen D. Fisher.

Find a linear fractional transformation $T$ that maps the real axis onto itself and the line $y=x$ onto the circle $|w+i|=\sqrt2$

where the convention is that a complex number is written in the form $w=x+iy; \{x,y\} \in \mathbb{R}$.

The "usual" method described so far in the book and used in the first problems of the chapter all work by the "method of triples" where three numbers $(z_1, z_2, z_3)$ on the original set is sent to three points $(w_1, w_2, w_3)$ on the image, and the transformation $L(z)$ that does this is given by

$$T(z)=\frac{z-z_1}{z-z_3}\frac{z_2-z_3}{z_2-z_1} \\ S(w)=\frac{w-w_1}{w-w_3}\frac{w_2-w_3}{w_2-w_1}=\frac{aw+b}{cw+d} \\ S^{-1}(z)=\frac{-dz+b}{cz-a} \\ L(z)=S^{-1}(T(z))$$

Using this method I can find a transformation that sends the triple $(0, 1+i, \infty)$ → $(-1-2i, -1, 1)$ where the first triple is on the line and the last three are on the final circle. With these points, my calculations give the transformation

$$L(z)=\frac{z-3-i}{z+1-i}$$

which can be checked to send the first triple to the second. But it doesn't at the same time send the real axis to itself, and I don't know how to modify it to retain the first property and attain the second, or if I should work from different principles altogether.

The solution portion of the book just gives the answer $L(z)=\frac{z-1}{z+1}$, which looks short and simple enough, so I can't imagine that it should be overly complicated to find it either.

After reading the chapter over and over I ask my first question on the mathematics portion of SE.

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The real line and $y=x$ intersect at $0$ and $\infty$ (in the extended complex plane), so these intersection points have to map to the intersection points of the real line and your circle, i.e., $\pm 1$. However, your solution maps $0$ to a different point, and that is why it does not work.
$$\frac{2z-1}{2z+1}$$ also works. Therefore we must check the image of one more point on the line $y=x$. For example $1+i$ goes to $(\frac{1}{5}, \frac{7}{5})$ under $L(x)$, which is on the circle. Then we are sure that we have the correct answer. – PAD Oct 11 '12 at 15:01
@PantelisDamianou, Lukas Geyer: Thanks to both of you. Matching intersections was the key, as you could already see. In the method of "triples to triples" I got the correct answer directly by rearranging my matching points to $(0, \infty, 1+i)$ &rarr; $(-1, 1, 1-2i)$. I had done a similar exercise earlier, but "lucked out" in that the intersections were made to be matching points by chance. Thanks again, you really made this exercise make sense to me. – Daniel Andersson Oct 11 '12 at 15:46