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.

I am trying to show that elements of the general Möbius group generated by an affine transformation $f(z) = az+b$, the inversion map $f(z)=\frac{1}{z}$ and complex conjugation $f(z)=\overline{z}$, where $a,b \in \mathbb{C}$, are conformal maps.

So if I have a curve lying in the $x$-$y$ plane, then its image in the $u$-$v$ plane is another curve, and I want to show that the angles between the curves are preserved under the three transformations above. Because we speak of angles between curves, the problem reduces to proving that the angle between their tangents is preserved and hence to the simplest problem f showing that angles between straight lines are preserved under these maps.

Now let $X_1$ and $X_2$ be straight lines. Then under an affine transformation and complex conjugation it is not hard to show that the angle between $X_1$ and $X_2$, let's call it $angle(X_1 , X_2)$ is preserved (Under complex conjugation the sign of the angle is reversed however).

Now the inversion map is a bit more tricky. If my two lines in $x-y$ space pass through the $y$-axis at 1, viz. that $X_k$ has equation $\beta_k z + \overline{\beta_k z} + 1 = 0$ where $ k \in {0,1}$, then under inversion these straight lines get mapped to circles through the origin, and hence it is plain that the angles between the tangents of the circles at $(0,0)$ is preserved.

However, what if my lines $X_1$ and $X_2$ are such that under inversion $X_1$ maps to a line while $X_2$ maps to a circle? This is where I am getting stuck.

On the other hand, if I try to prove the conformality of the inversion map by saying that my $x$ and $y$ coordinate curves in $x-y$ space get mapped to $\frac{x}{x^2 + y^2}$ and $\frac{-y}{x^2 + y^2}$ in $u-v$ space, and then write vector equations of lines and compute partial derivates, the maths becomes ugly and the method is certainly not elegant.

Anyone got any ideas? (Maybe involving Cauchy - Riemann Equations!!)

share|improve this question
    
Try using the following definition, which for example can be found in Rudins "real and complex analysis": $f: \Omega \to \mathbb C$ is conformal at $z_0 \in \Omega$, iff the map $z \mapsto f(z)-f(z_0)$ has an isolated zero at $z_0$ and for every $\theta \in [0,2\pi)$ the limit $\lim_{r\to 0, r >0} e^{-i\theta}\frac{f(z_0+re^{i\theta}) - f(z_0)}{|f(z_0+re^{i\theta}) - f(z_0)|}$ exists and is independent of $\theta$. You can easily see that if $f$ is differentiable at $z_0$ and $f'(z_0)\not = 0$ then it is also conformal at $z_0$. –  Alexander Thumm Mar 28 '11 at 10:41
    
So you say the maths is not elegant and becomes ugly, but did you work through it? It is the most direct and elementary way to see the conformality of the inversion map in my opinion. –  Glen Wheeler Mar 28 '11 at 12:06

3 Answers 3

There is a geometric way of seeing that inversion works properly, which is based on considering not $1/z$ but $\bar{1/z}$. Then your $(x,y)$ gets mapped to $(\frac x{x^2+y^2},\frac y{x^2+y^2})$, which can be interpreted geometrically as follows. Let $O$ be the origin, and let $P$ be a point. Then $P$ gets sent to the unique point $P'$ on the ray $\vec{OP}$ such that $OP\cdot OP'=1$ (where $OP$ is the length of the segment joining $O$ and $P$). If $R$ is any point on the unit circle between $P$ and $P'$, then we can rewrite the equation as $\frac{OP}{OR}=\frac{OR}{OP'}$. Hence, triangles $OPR$ and $ORP'$ will be similar. You can then prove that triangles $OPQ$ and $OQ'P'$ are similar for any points $P$ and $Q$. (multiply to get $\frac{OP}{OR}\frac{OR}{OQ}=\frac{OQ'}{OR}\frac{OR}{OP'}$).

Once you have that it is easy to show that if a line $\ell$ does not go through the origin, it gets mapped to a circle that goes through the origin as follows. Suppose that $P$ is on $\ell$ such that $\vec{OP}$ is perpendicular to $\ell$. Then for any other point $Q$ on $\ell$ we have that triangle $OPQ$ has right angle $OPQ$, and hence triangle $OQ'P'$ has right angle $OQ'P'$ for all points $Q'$. But that implies that the points $Q'$ trace out a circle with diameter $OP'$. Hence, $\vec{OP}$, which was perpendicular to $\ell$ is perpendicular to the circle that $\ell$ got sent to. This shows full conformality since if you have two lines $\ell_1$ and $\ell_2$, you have three cases.

  1. $\ell_1$ and $\ell_2$ go through the origin -- they are fixed by inversion.
  2. $\ell_1$ goes thorugh the origin, but $\ell_2$ does not. Then $\ell_2'$ which is the perpenciular line to $\ell_2$ though the origin is fixed along with $\ell_1$ and perpendicular to the image-circle of $\ell_2$, hence the angle between $\ell_1$ and the circle equals the angle between $\ell_1$ and $\ell_2$.
  3. Both don't go through the origin, in which case take both of their perpendiculars $\ell_1$ and $\ell_2$.

There is also the analytic way of doing this.

If we consider $\mathbb C$ as the real plane $\mathbb R^2$, then what you're trying to show is that certain functions $f\colon \mathbb R^2\to\mathbb R^2$ are angle-preserving.

To be angle preserving at a particular point means that the derivative $df(z_0)$, which is a $2\times 2$ real matrix, preserves the angles between vectors. To see this, take two (smooth) curves $c_1\colon\mathbb R\to\mathbb R^2$ and $c_2\colon\mathbb R\to\mathbb R^2$. The angle between them at point $z_0=c(t_1)=c(t_2)$ is, as you said, the angle between the tangent lines, which is of course the angle between the tangent vectors $dc_1(t_1)$ and $dc_2(t_2)$ -- let's call them $v_1$ and $v_2$.

The function $f$ will map the two curves to the curves $f\circ c_1$ and $f\circ c_2$, whose tangent vectors will be $df(z_0)v_1$ and $df(z_0)v_2$. Thus, we want $df(z_0)$ to be a matrix $M$ that preserves the angles between vectors.

Fortunately, you're not asking how to classify all conformal mappings, so we just need to check that the derivatives of $z\to az+b$, $z\to 1/z$ and $z\to\bar z$ preserve angles at every $z_0$.

It turns out that the Cauchy-Riemann equations tell you that complex-differentiable functions (such as $z\to az+b$ and $z\to 1/z$) have derivates that look like $M=\left[\begin{matrix}a&b\\-b&a\\\end{matrix}\right]$ which since $\det M=a^2+b^2$ look exactly like $a^2+b^2\left[\begin{matrix}\cos(\theta)&\sin(\theta)\\-\sin(\theta)&\cos(\theta)\end{matrix}\right]$ for some angle $\theta$, which is exactly dilation by a factor of $a^2+b^2$ and rotation by angle $\theta$, which evidently preserve angles between vectors (for a quick proof, the rotation matrix is orthonormal and so actually preserves dot products, while dilation merely scales dot products by some constant).

To get the action of $\bar z$, note that it sends $x+iy$ to $x-iy$ and so its derivative is the matrix $\left[\begin{matrix}1&0\\0&-1\end{matrix}\right]$. This matrix does not preserve the dot product -- a short computation shows that it reverses the sign of the dot product. But that is ok since reversing the sign of the dot product is a reversal of orientation, i.e. the angle changes from $\theta$ to $180^o-\theta$, which for the purposes of being conformal is the same thing.

share|improve this answer
    
Dear @VladimirSotirov, I know what you mean by "[the matrix of complex conjugation $M$] does not preserve the dot product" but I think you didn't word it very well. In fact, this matrix is orthonormal, so it preserves the dot product, although it does not preserve the orientation. I think that's what you meant by "reverses the sign of the dot product", but I think such terminology is unclear because the dot product changes by $M$ by a congruence transformation, and again because $M$ is orthonormal, the dot product doesn't change. –  Rodrigo Mar 30 at 15:04

Unfortunately I am not up to the level of understanding complex analysis (I am first year student). Is there any way simpler to get around this?

I'll give it a shot: I know that lines that don't pass through the origin in $\mathbb{C}$ get mapped to a circle through the origin, because:

Let $X$ be an euclidean line not through the origin in $\mathbb{C}$. Then $X$ has equation

$\beta z + \overline{\beta z} + 1 = 0$, for some $\beta$ a complex number. Because of the presence of a $1$, when I let $w=\frac{1}{z}$ I will always have the term $w\overline{w}$ and so such a line not through the origin is an Euclidean circle in $\mathbb{C}$.

Specifically it is an euclidean circle that passes through the origin as I can write the equation $w\overline{w} + \overline{\beta}z + \overline{z} \beta = 0$ into $|w+\beta|^2 = |\beta|^2$.

Oh I think I get it now. (Should not rejoice so quickly). Say for example I have a two lines, one of which passes through the origin, say $X_1$ in $\mathbb{C}$ while the other, say $X_2$ does not. Then $X_1$ maps to a line through the origin under inversion, and $X_2$ maps to a circle through the origin as shown above in this reply. Then to prove conformality does it suffice just to show that:

The angle between the lines $X_1$ and $X_2$ is the same as the angle of the tangent to the circle, $f(X_2)$ where $f(z) = \frac{1}{z}$ at the origin and the line $f(X_1)$?

If my line $f(X_2)$ (the circle) $f(x_1)$ at another point, must I consider the tangent the other point? E.g. as specific examples take:

$X_1 = \{z\in\mathbb{C} : z(1-i) + \overline{z}(1+i) = 0\}$ and $X_1 = \{z\in\mathbb{C} : \overline{z}(1-i) - z(1+i) +2 = 0\}$,

hence

$f(X_1) = \{z \in \mathbb{C} : \overline{z}(1-i) + z(1+i) = 0\}$ and $f(X_2) = \{z \in \mathbb{C} : z(i-1) - \overline{z}(1+i) + 2z\overline{z}= 0\}$

where $f(z)=\frac{1}{z}$. We find that this line and circle intersect at points $(0,0)$ and $(\frac{2^{\frac{1}{4}} + 1}{2},\frac{2^{\frac{1}{4}} + 1}{2})$.

share|improve this answer
    
@ GlenWheeler, let's say we go with this direct approach: Let $\vec{r_{1}}=\left(\begin{array}{c} a_{1}\\ b_{1}\end{array}\right)+t\left(\begin{array}{c} x_{1}\\ y_{1}\end{array}\right)$ and $\vec{r_{2}}=\left(\begin{array}{c} a_{2}\\ b_{2}\end{array}\right)+t\left(\begin{array}{c} x_{2}\\ y_{2}\end{array}\right)$ be vectors in the $x-y$ plane. Then I would first need to substitute these into the expressions in terms of $x$ and $y$ in $u-v$ space and calculate $\frac{\partial\vec{r_{1}}}{\partial t}$ and similarly for the $\vec{r_{2}}$. I have expressions with 30 terms in Mathematica! –  user38268 Mar 28 '11 at 13:06
    
Benjamin Lim (To get my attention, try putting my name directly after the 'at' symbol like this: @Glen Wheeler) 30 terms seems a little extreme. Compute the angle between the two lines, invert the lines and compute the new angle. There are plenty of slick ways to do this (like writing the lines in a different coordinate system) but it is still tractable with the plain old coordinate system in $\mathbb{R}^2$. –  Glen Wheeler Mar 30 '11 at 8:58

I'm not sure I understand exactly what your problem is, but maybe Section 3.II.4 (pp. 130–132) of Needham's Visual Complex Analysis helps. (Google Books link.)

share|improve this answer

Your Answer

 
discard

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