Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

1 Answer 1

up vote 1 down vote accepted

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.

share|improve this answer
    
$$\frac{2z-1}{2z+1}$$ 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 $(0, \infty, 1+i)$ → $(-1, 1, 1-2i)$. 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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.