A conformal mapping is a map $f:U\to V$ with $U,V\subseteq\mathbb{C}$ such that the angles are locally preserved. This can be reformulated saying the jacobian matrix is everywhere a scalar multiple of a rotation matrix. Note: I am considering only the case that is relevant to my problem. For a more general definition of conformal mapping, see Wikipedia. More precisely, a map is conformal at a point $u$ if for any two curves intersecting at $u$ the angle they form at $u$ is the same as that their images form at $f(u)$. Now I have been told that $f(z)=z^{\frac{3}{4}}$, as well as any power function, is conformal. But then take the two lines connecting the origin to $e^{-i\pi\frac23}$ and $e^{i\pi\frac23}$. They are mapped to the two semilines of the vertical axis. Their angle was $\frac43\pi$ and has become $\pi$. So how can this $f$ be conformal? It is definitely not conformal at the origin, right? So where is it conformal?

Edit: Taking a closer look at the article, I saw that a holomorphic function is conformal if and only if its derivative is nonzero on all $U$. This means that the function $f$ is conformal outside the origin, since it is holomorphic. So any function that is holomorphic is conformal, I mean locally. Is that right and how do I prove it?

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    $\begingroup$ write down the derivative of the holomorphic map as a matrix and use the Cauchy Riemann equations, you will then see it has the form you mentioned (scalar multiple of a rotation matrix). $\endgroup$ – Thomas Oct 30 '14 at 16:22
  • $\begingroup$ Let's see. The Cauchy-Riemann equations say that, if $f(x,y)=(u,v)$ seen as $f(x+iy)=u+iv$ is holomorphic, then $\partial_xu=\partial_yv$ and $\partial_yu=-\partial_xv$. Then the Jacobian matrix will be: $$\begin{array}{cc} \partial_xu & \partial_yu \\ \partial_xv & \partial_yv \end{array}=\begin{array}{cc} \partial_xu & -\partial_xv \\ \partial_xv & \partial_xu \end{array}=\begin{array}{cc} a & -b \\ b & a \end{array}.$$ Now if we place $\sqrt{a^2+b^2}$ in front of the matrix and divide its entries by it, we get a rotation matrix. This means the Cauchy-Riemann equations imply conformality. $\endgroup$ – MickG Oct 30 '14 at 16:31
  • $\begingroup$ It is clear that the converse holds, so conformality at a point is equivalent to holomorphicity (or rather to the C-R equations which are equivalent to holomorphicity) at that point. This means that all powers, where they are holomorphic, are conformal. NOw I missed out that $\sqrt{a^2+b^2}$ may be 0, which makes my implication not valid since division by 0 is impossible outside limits. So where the derivatives are 0 (and thus $f'(z)=0$) there is no conformality. $\endgroup$ – MickG Oct 30 '14 at 16:37

$\newcommand{\pd}{\partial} \newcommand{\fr}{\dfrac} \newcommand{\pDer}[2][]{\fr{\pd #1}{\pd #2}}$ I post this to get the question off the unanswered list.

First of all, the example $f$ is conformal everywhere save for the origin. That is because its derivative is $f'(z)=\frac34z^{-\frac14}$, which is nonzero and well-defined everywhere outside the origin, unless we want to include infinity points into $\mathbb{C}$, where the function would be 0.

As I proved in the comments based on Thomas's suggestion, holomorphicity with nonzero derivative is equivalent to conformality (locally, I mean, i.e. at any given point $p\in\mathbb{C}$). I'll copy the proof here. Conformality implies the Jacobian of the function when viewed as $f:\mathbb{R}^2\to\mathbb{R}^2$ is a scalar multiple of a rotation matrix. Let us then write the Jacobian of any function. First of all we identify $x+iy$ and $(x,y)$, and all similar complex number -- point on the plane pairs. With that, the Jacobian of $f$ will be: $$\mathrm{Jac}_f(P)=\left(\begin{array}{cc} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y}\end{array}\right),$$ where $f(x,y)=(u,v)$ and the derivatives are all evaluated at point $P=(x_0,y_0)$. By the Cauchy-Riemann equations, which characterize holomorphicity (i.e. are equivalent to it), the Jacobian becomes: $$\mathrm{Jac}_f(P)=\left(\begin{array}{cc} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ -\frac{\partial u}{\partial y} & \frac{\partial u}{\partial x} \end{array}\right).$$ If either of those two derivatives is nonzero, which is equivalent to $f'(P)=0$ when $f$ is viewed as $f(x+iy)=u+iv$, then $\sqrt{(\frac{\partial u}{\partial x})^2+(\frac{\partial u}{\partial y})^2}$ is also nonzero, so dividing the entries of the matrix by it and putting it in front of the resulting matrix as a scalar multiplicator we get, by calling the root $\alpha$ and $a=\frac{1}{\alpha}\frac{\partial u}{\partial x}$ and $b=-\frac{1}{\alpha}\frac{\partial u}{\partial y}$: $$\mathrm{Jac}_f(P)=\alpha\left(\begin{array}{cc} a & -b \\ b & a \end{array}\right).$$ Since the derivatives are not both zero, it follows that one of $a,b$ (or both) is nonzero. Also, by definition of $a,b$ we have $a^2+b^2=1$, and both are in $[0,1]$. This means they are the sine and cosine of some angle. Substituting $a=\cos\theta,\,b=\sin\theta$, the Jacobian becomes the multiple by $\alpha$ of the rotation by angle $\theta$, so holomorphicity with nonzero derivative implies conformality. It is evident that the passages I made are invertible, so holomorphicity with nonzero derivative is in fact equivalent to conformality.

Of course, if the function has zero derivative or is not holomorphic, we have problems. In particular, if it is antiholomorphic, its conjugate is holomorphic and therefore conformal, if the derivative is nonzero. If one were to derive the antiholomorphic equivalent of the Cauchy-Riemann equations, and plug them into the Jacobian, one would probably find a reflection matrix. This means angles are preserved, but only in magnitude, and inverted in orientation, as Wikipedia states.

Evidently, $f$ is not conformal at the origin, but then neither is it holomorphic, so no problem. Indeed, since we have proved an equivalence, we can deduce any power function is not conformal at the origin, save for the trivial case of the identity, since save for that case any power function is either not holomorphic in the origin or has derivative zero. That is because $f(z)=z^\alpha\implies f'(z)=\alpha z^{\alpha-1}$, and if $\alpha>1$ that yields a power with positive exponent which is 0 in the origin, if $\alpha<1$ this yields a power with negative exponent which diverges in the origin, and for $\alpha=1$ this yields the trivial case of the identity. This can be taken to a far more general level if we assume that any function which is differentiable when the variable is real is holomorphic with the same derivative in the complex case -- e.g., that the logarithm $\log x$ is holomorphic and has derivative $\frac{1}{x}$. In that case, any such function is conformal where it is differentiable. That would mean, for example, that the exponential is everywhere holomorphic, and that the logarithm is holomorphic everywhere outside the origin. A similar assumption was in fact made with powers, since I have only sketched the proof of that for natural (i.e. $\mathbb N$) exponents.

To be precise, what we proved is that for holomorphic functions, nonzero derivative is equivalent to conformality. Of course, if the function is conformal it must be holomorphic, since the Jacobian has the form of a multiple of a rotation and that implies the Cauchy-Riemann equations and therefore holomorphicity.

Extra: To avoid asking a new question for a marginally related thing, I will try to show that if $\phi:\mathbb{R}^2\to\mathbb{R}$ satisfies $\nabla^2\phi=0$, i.e. is harmonic, and $f$ is conformal, then $\phi\circ f$ is still harmonic. I guess we just need to calculate the laplacian: $$\nabla^2(\phi\circ f)=\frac{\partial^2\phi\circ f}{\partial x^2}+\frac{\partial^2f\circ\phi}{\partial y^2}=$$ $$=\fr{\pd}{\pd x}\left(\pDer[\phi]{f_1}\pDer[f_1]{x}+\pDer[\phi]{f_2}\pDer[f_2]{x}\right)+\fr{\pd}{\pd y}\left(\pDer[\phi]{f_1}\pDer[f_1]{y}+\pDer[\phi]{f_2}\pDer[f_2]{y}\right)=$$ $$=\pDer[\phi]{f_1}\pDer[^2f_1]{x^2}+\pDer[f_1]{x}\pDer{x}\pDer[\phi]{f_1}+\pDer[\phi]{f_2}\pDer[^2f_2]{x^2}+\pDer[f_2]{x}\pDer{x}\pDer[\phi]{f_2}+$$ $$+\pDer[\phi]{f_1}\pDer[^2f_1]{y^2}+\pDer[f_1]{y}\pDer{y}\pDer[\phi]{f_1}+\pDer[\phi]{f_2}\pDer[^2f_2]{y^2}+\pDer[f_2]{y}\pDer{y}\pDer[\phi]{f_2}=$$ $$=\pDer[\phi]{f_1}\nabla^2f_1+\pDer[\phi]{f_2}\nabla^2f_2+\pDer[f_1]{x}\pDer{x}\pDer[\phi]{f_1}+\pDer[f_2]{x}\pDer{x}\pDer[\phi]{f_2}+\pDer[f_1]{y}\pDer{y}\pDer[\phi]{f_1}+\pDer[f_2]{y}\pDer{y}\pDer[\phi]{f_2}.$$ Now the laplacians of $f_1$ and $f_2$ are 0, since the real and imaginary part of a holomorphic function (like $f$, which is conformal and therefore holomorphic) are always harmonic, as can be easily proved by writing the laplacian and plugging in the Cauchy-Riemann equations. So we are left with the rest: $$\nabla^2(\phi\circ f)=\left(\pDer[f_1]{x}\right)^2\pDer[^2\phi]{f_1^2}+\left(\pDer[f_2]{x}\right)^2\pDer[^2\phi]{f_2^2}+\left(\pDer[f_1]{y}\right)^2\pDer[^2\phi]{f_1^2}+\left(\pDer[f_2]{y}\right)^2\pDer[^2\phi]{f_2^2}.$$ Now what is the laplacian of $\phi$ we have supposed to be 0? $\phi$ was harmonic before the composition, so its laplacian is 0 in terms of the old coordinates, which are now $f_1,f_2$: $$\nabla^2\phi=\pDer[^2\phi]{f_1^2}+\pDer[^2\phi]{f_2^2}=0.$$ Keeping this in mind, we rewrite the terms above "wisely", getting: $$\nabla^2(\phi\circ f)=\left(\pDer[f_1]{x}\right)^2\pDer[^2\phi]{f_1^2}-\left(\pDer[f_2]{x}\right)^2\pDer[^2\phi]{f_1^2}+\left(\pDer[f_1]{y}\right)^2\pDer[^2\phi]{f_1^2}-\left(\pDer[f_2]{y}\right)^2\pDer[^2\phi]{f_1^2}.$$ It seems then that it is enough for $f$ to be holomorphic for this to vanish. INdeed, the Cauchy-Riemann equations read: $$\left\lbrace\begin{array}{@{}l@{}} \pDer[f_1]{x}=\pDer[f_2]{y} \\ \pDer[f_1]{y}=-\pDer[f_2]{x} \end{array}\right.$$ Plugging this into the above, we can easily see they imply the laplacian we wanted to vanish does in fact vanish, ending this infinite series of calculation and proving the following:

If $\phi:\mathbb{R}^2\to\mathbb{R}$ is harmonic and $f:\mathbb{R}^2\to\mathbb{R}^2$ is holomorphic (where $\mathbb{R}^2$ and $\mathbb{C}$ are identified in the usual way $(x,y)\mapsto x+iy$), then $\phi\circ f$ is still harmonic.

More precisely, if $\phi$ is harmonic over a region $U$ and $f$ is holomorphic over a region $V$, then $\phi\circ f$ will be harmonic over $f^{-1}(U)\cap V$.

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  • $\begingroup$ Actually (to give you some more food for thought ;-), $f \mathbb{R}^2\rightarrow \mathbb{R}$ is harmonic iff it is locally the real or imaginary part of a holomorphic function. In particular real and imaginary part of a holomorphic function are both harmonic, they are called conjugate harmonic functions and if one is given the other one is uniquely determined up to a constant. And it's gradient will vanish iff any of corresponing harmonic function has nonzero complex derivative. If you know that, your extra follows automatically. $\endgroup$ – Thomas Oct 31 '14 at 5:31
  • $\begingroup$ OK so if $\varphi$ is harmonic one can easily find another function $\varphi'$ that satisfies the CR equations, thus making $\varphi+i\varphi'$ holomorphic: one merely needs to compose with $(x,y)\mapsto(y,-x)$. Obviously, shifting by a constant doesn't alter the derivatives. "its gradient" whose? $\endgroup$ – MickG Oct 31 '14 at 6:19
  • $\begingroup$ 'one merely needs to compose' -- that won't do, it's not that easy ;-). The gradient of your $\varphi$. $\endgroup$ – Thomas Oct 31 '14 at 10:29
  • $\begingroup$ Have a look at this page: en.wikipedia.org/wiki/Harmonic_conjugate $\endgroup$ – Thomas Oct 31 '14 at 10:35
  • $\begingroup$ Let $\phi'(x,y)=\phi(y,-x)$. Then $\frac{\partial\phi'}{\partial x}=\frac{\partial\phi}{\partial x'}\frac{\partial x'}{\partial x}+\frac{\partial\phi}{\partial y'}\frac{\partial y'}{\partial x}=\frac{\partial\phi}{\partial x'}\cdot0+\frac{\partial\phi}{\partial y'}\cdot1=\frac{\partial\phi}{\partial y'}$, $(x',y')$ being the coordinates where $\phi$ is evalued, making $\frac{\partial\phi}{\partial y'}$ none other than $\frac{\partial\phi}{\partial x}$, and similarly $\frac{\partial\phi'}{\partial y}=-\frac{\partial\phi}{\partial x}$, so CR is satisfied and $\phi+i\phi'$ is holomorphic. $\endgroup$ – MickG Oct 31 '14 at 12:58

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