2

I'm trying find the image of the disc $D(i,1)$ under the transformation $f(z) = \displaystyle\frac{3z-2}{2z+i}$ but i find no symmetries between the points of $D$ and $f(D)$, I notice that if $z \in D$ then $-\bar{z} \in D$ but this doesn't mean that $f(-\overline{z}) = -\overline{f(z)}$ so i can't find two symmetric points on $f(D)$ to compute the center and radii of the circunference.

I'm trying this method of the simetric point because i saw that if the circunference has center on the real line all mobius transformation has the simetry $f(\overline{z}) = \overline{f(z)}$ so its easy to find two points on the image and compute it center.

Has anyone a hint about this method or other way to solve this question?

InsideOut
  • 7,003
Eduardo Silva
  • 1,425

3 Answers3

1

The disk $D(\mathrm i,1)$ is given by $z \in \mathbb C$ for which $|z-\mathrm i| \le 1$.

Let $w$ be the coordinate on the image, i.e. $f : \mathbb C_z \to \mathbb C_w$ where $$w = \frac{3z-2}{2z+\mathrm i}$$ Rearranging to make $z$ the subject gives $$z = \frac{\mathrm i w + 2}{3-2w}$$ and in turn $$z-\mathrm i = \frac{3\mathrm i w+(2-3\mathrm i)}{2w-3}$$ Given that $|z-\mathrm i| \le 1$ we see that $$\left| \frac{3\mathrm i w+(2-3\mathrm i)}{2w-3} \right| \le 1 \implies |3\mathrm i w+(2-3\mathrm i)| \le |2w-3|$$ Since $|-\mathrm i\cdot z| = |-\mathrm i|\cdot |z|=1\cdot |z|=|z|$ for all $z \in \mathbb C$ we see that $$|3\mathrm i w+(2-3\mathrm i)|=|3 w-(3+2\mathrm i)|$$ It follows that $|3 w-(3+2\mathrm i)| \le |2w-3|$.

If we put $w=u+\mathrm iv$ then we have \begin{eqnarray*} |3 (u+\mathrm iv)-(3+2\mathrm i)| &\le& |2(u+\mathrm iv)-3| \\ \\ |(3u-3)+\mathrm i(3v-2)| &\le& |(2u-3)+\mathrm i(2v)| \\ \\ \sqrt{(3u-3)^2 + (3v-2)^2} &\le& \sqrt{(2u-3)^2+(2v)^2} \\ \\ {(3u-3)^2 + (3v-2)^2} &\le& {(2u-3)^2+(2v)^2} \\ \\ 5u^2+5v^2-6u-12v+4 &\le& 0 \\ \\ \left(u-\frac{3}{5}\right)^{\! 2} + \left(v-\frac{6}{5}\right)^{\! 2} &\le& 1 \end{eqnarray*} The image of $D(\mathrm i,1)$ is $D(\frac{3}{5}+\frac{6}{5}\mathrm i,1)$.

Fly by Night
  • 32,886
  • it's a coincidence or the radius will always be preserved? – Eduardo Silva Dec 09 '16 at 17:18
  • there is a geometric way to solve this question? – Eduardo Silva Dec 09 '16 at 17:21
  • @EduardoSilva It's a coincidence. Something as simple as $f(z)=2z$ will change the radius. Mobius transformations can send circles or circles to lines (finite radius to infinite radius). They can also turn circles inside out. So the inside of a circle is sent to the outside of another circle. – Fly by Night Dec 09 '16 at 17:21
  • looking to the points on the boundary – Eduardo Silva Dec 09 '16 at 17:23
  • @EduardoSilva I can't see how looking at the boundary helps. It's the same method to find the image of the boundary circle. Please remember to up-vote answers that you find useful; it's always nice to say thank you :) – Fly by Night Dec 09 '16 at 18:09
0

If you are lazy you can pick $3$ points on the circumference of the circle, compute their image by $f$, and then you obtain $3$ points on the circumference of the image of the circle, which is enough to determine the circle.

Another algebraic way to compute the result is that the equation of a circle (or a line if the coefficient of $z\bar z$ is $0$) is something of the form $z\bar z + az + \bar a \bar z +b = 0$ where $a$ is complex and $b$ is real, or equivalently, it is a relation $\bar z = (-az-b)/(z+\bar a) = g(z)$ for some Mobius transform $g$ satisfying $\bar g = g^{-1}$.

And so to translate this relation between $z$ and its conjugate to get a relation between $f(z)$ and its conjugate, you simply write $\overline {f(z)} = \bar f (\bar z) = (\bar f \circ g) (z) = (\bar f \circ g \circ f^{-1}) (f(z))$

And so the Mobius transform used to describe the image of the new circle is $h = \bar f \circ g \circ f^{-1}$. (And $h^{-1} = \bar h$ still holds so it gives you a circle or a line)

In any case, you have to do a bunch of computation.

Here the equation of the circle is $(z-i)(\bar z+i) = 1$, so $z\bar z + iz - i\bar z = 0$, which can be written as $\bar z = g(z)$ with $g(z) = -iz/(z-i)$.
Meanwhile, $f^{-1}(w) = (iw+2)/(3-2w)$, so
$(g \circ f^{-1})(w) = -i(iw+2)/((iw+2)-i(3-2w)) = (w-2i)/(3iw+2-3i)$,
then $(\bar f \circ g \circ f^{-1}) (w) = (3(w-2i)-2(3iw+2-3i))/(2(w-2i)-i(3iw+2-3i)) = ((3-6i)w-4)/(5w-(3+6i))$

Then the circle is given by the equation $5w\bar w - (3+6i)\bar w - (3-6i)w + 4 = 0$, which means after multiplying by $5$ and factoring, you get $|5w - (3+6i)|^2= |3+6i|^2-20 = 45-20 = 25 = 5^2$. So this is the circle with center $\frac 15(3+6i)$ and radius $1$

mercio
  • 51,119
0

The method I came up with for this answer says that for the transform $\frac{3z-2}{2z+i}$ and the circle with radius $1$ centered at $i$, we need to map the antipodal points $$ i\pm\frac{i+i/2}{|i+i/2|}\cdot1=\{2i,0\} $$ through the transform to get the antipodal points in the image $$ \left\{\frac65+\frac25i\,,\,2i\right\} $$ The radius of the image circle is $\frac{\left|\frac65+\frac25i-2i\right|}2=1$ and its center is $\frac{\frac65+\frac25i+2i}2=\frac35+\frac65i$.

robjohn
  • 353,833