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How to compute a linear fractional transformation that maps a circle to a given circle? For example, let $C_1$ be the circle $|z+(2+i)|=1$ and $C_2$ be the circle $|z-5|=7$. How to find a linear fractional transformation that maps $C_1$ to $C_2$? Thank you very much.

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3 Answers 3

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As I said in your previous question, a LFT is uniquely determined by 3 points and their images. If $z_1, z_2, z_3$ are your inputs (pick 3 points on your first circle) and $w_1, w_2, w_3$ are your images of these (pick 3 points on your second circle). Be sure they are in the same order as you move around the circle, as I mentioned before that LFTs preserve order. Then, use the formula $$\frac{(z_1 - z_3)(z_2 - z)}{(z_2 - z_3)(z_1 - z)} = \frac{(w_1 - w_3)(w_2 - w)}{(w_2 - w_3)(w_1 - w)}$$ and then solve for $w$ to get $w$ as a function of $z$. This formula is one you need to know and is useful in many, many situations, not just this one.

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To map a circle on a circle, you only need to translate and to scale (no fractional transformation is necessary).

If the first circle is centered at $z_1$ with radius $R_1$ and the second circle is at $z_2$ with radius $R_2$ correspondingly, then $$ z \mapsto z- z_1$$ translates the first circle to the origin. Then $$ z \mapsto \frac{R_2}{R_1} z$$ rescales the radius. Finally, $$z \mapsto z+ z_2$$ translates the circle to the origin of the second (image) circle. Composing these transformations yields the mapping $$f(z) = \frac{R_2}{R_1} (z-z_1) + z_2$$ which is actually a linear transformation.

Comment: for your example $z_1 = -2 - i$, $R_1=1$, $z_2 = 5$, $R_2=7$, and $$f(z) = \frac{1}{7} [ z + (2+i)] + 5.$$

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$+1$ for such a geometric explanation –  Une Femme Douce Apr 27 '13 at 22:05

Choose thre points $\{z_1,z_2,z_3\}$ in $C_1$ and three points $\{w_1,w_2,w_3\}$ in $C_2$. Then find the linear fractional transformation $f$ such that $f(z_i)=w_i$, $1\le i\le3$. Some care must be taken with the order of the points. Choose them in such a way that when you travel through them, the interior of the circle is on the left side.

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