Conformal Transformations and Singularities

The validity of the extended version of the Cauchy-Goursat theorem can be generalized to all complex analytic functions integrated on arbitrarily given closed integration contours, through the use of conformal transformations. In order to do this, let us first establish the definition and the notation for a conformal transformation. This is essentially a shorter version of the discussion on this topic which was given in Section 4 of a previous paper [#!CAoRFV!#]. Consider therefore two complex variables $z_{a}$ and $z_{b}$ and the corresponding complex planes, a complex analytic function $\gamma(z)$ defined on the complex plane $z_{a}$ with values on the complex plane $z_{b}$, and its inverse function, which is a complex analytic function $\gamma^{(-1)}(z)$ defined on the complex plane $z_{b}$ with values on the complex plane $z_{a}$,

$$z_{b} = \gamma(z_{a}),$$

$$z_{a} = \gamma^{(-1)}(z_{b}).$$ (13)

Let us point out here that these relations immediately imply that

$$\frac{dz_{b}}{dz_{a}} = \frac{d\gamma(z_{a})}{dz_{a}},$$

$$\frac{dz_{a}}{dz_{b}} = \frac{d\gamma^{(-1)}(z_{b})}{dz_{b}},$$ (14)

which in turn imply that

$$\frac{d\gamma(z_{a})}{dz_{a}}\, \frac{d\gamma^{(-1)}(z_{b})}{dz_{b}} = 1,$$ (15)

for all pairs of points $z_{a}$ and $z_{b}$ related by the conformal transformation. This means that any point where the derivative of $\gamma(z_{a})$ has a zero on the $z_{a}$ plane corresponds to a point where the derivative of $\gamma^{(-1)}(z_{b})$ has a singularity on the $z_{b}$ plane, and vice-versa.

Consider a bounded and simply connected open region $S_{a}$ on the complex plane $z_{a}$ and its image $S_{b}$ under $\gamma(z)$, which is a similar region on the complex plane $z_{b}$. It can be shown that if $\gamma(z)$ is analytic on $S_{a}$, is invertible there, and its derivative has no zeros there, then its inverse function $\gamma^{(-1)}(z)$ has these same three properties on $S_{b}$, and the mapping between the two complex planes established by $\gamma(z)$ and $\gamma^{(-1)}(z)$ is conformal, in the sense that it preserves the angles between oriented curves at points where they cross each other. The famous Riemann mapping theorem states that such a conformal transformation $\gamma(z)$ always exists between the open unit disk $S_{a}$ and any region $S_{b}$. In addition to this, these properties of $\gamma(z)$ and $\gamma^{(-1)}(z)$ can be extended to the boundary of the regions so long as these boundaries are differentiable simple curves.

Consider therefore that the regions under consideration are the interiors of simple closed curves. One of these curves will be the unit circle $C_{a}$ on the complex plane $z_{a}$, and the other will be a given closed differentiable simple curve $C_{b}$ on the complex plane $z_{b}$. Since $\gamma(z_{a})$, being analytic, is in particular a continuous function, the image on the $z_{b}$ plane of the unit circle $C_{a}$ on the $z_{a}$ plane must be a continuous closed curve $C_{b}$. We can also see that $C_{b}$ must be a simple curve, because the fact that $\gamma(z_{a})$ is invertible on $C_{a}$ means that it cannot have the same value at two different points of $C_{a}$, and therefore no two points of $C_{b}$ can be the same. Consequently, the curve $C_{b}$ cannot self-intersect. Finally, the fact that $C_{b}$ must be a differentiable curve is a simple consequence of the facts that the $\gamma(z_{a})$ transformation is conformal and that the unit circle $C_{a}$ is a differentiable curve.

Given any analytic function $w_{a}(z_{a})$ on the $z_{a}$ plane, the conformal transformation $\gamma(z_{a})$ maps it to a corresponding function $w_{b}(z_{b})$ on the $z_{b}$ plane, and the inverse conformal transformation $\gamma^{(-1)}(z_{b})$ maps that function back to the function $w_{a}(z_{a})$ on the $z_{a}$ plane. We can do this by simply composing either $w_{b}(z_{b})$ or $w_{a}(z_{a})$ with either the transformation or its inverse, and simply passing the values of the functions,

$$w_{b}(z_{b}) = w_{a}(z_{a})$$

$$= w_{a}\!\left(\gamma^{(-1)}(z_{b})\right),$$

$$w_{a}(z_{a}) = w_{b}(z_{b})$$

$$= w_{b}\!\left(\gamma(z_{a})\right).$$ (16)

Since the composition of two analytic functions is also an analytic function, and since $\gamma(z_{a})$ is analytic on the closed unit disk, whenever $w_{b}(z_{b})$ is analytic on the $z_{b}$ plane the corresponding function $w_{a}(z_{a})$ will also be analytic at the corresponding points on the $z_{a}$ plane. Of course, where one of these two functions has an isolated singularity on its plane of definition, so will the other on the corresponding point in the other plane. Note that if any of these functions is integrated over a closed integration contour on which it has any isolated integrable singularities, whenever these singularities are branch points we assume that the corresponding branch cuts extend outward from the integration contours.

The concepts of a soft singularity and of a borderline hard singularity can be immediately extended from the case of inner analytic functions within the unit disk to singularities of arbitrary complex analytic functions anywhere on the complex plane. The concept of a soft singularity of $w(z)$ at $z_{1}$ depends only on the existence of the $z\to z_{1}$ limit of $w(z)$. The concept of a hard singularity of $w(z)$ at $z_{1}$ depends only on the non-existence of that same limit. Finally, the concept of a borderline hard singularity can be defined as that of a hard singularity which is nevertheless an integrable one. In order to discuss what happens with the singularities under the conformal transformation, we must establish a few simple preliminary results, by means of the following lemmas.

Lemma 1   : If $z_{b,1}$ is a singular point of $w_{b}(z_{b})$, then the corresponding point $z_{a,1}$ under the conformal transformation is a singular point of $w_{a}(z_{a})$.

Since according to Equations (16) we have that $w_{b}(z_{b,1})=w_{a}\!\left(\hspace{-1.4pt}\gamma^{(-1)}(z_{b,1})\right)$ and since $\gamma^{(-1)}(z_{b})$ is analytic within and on the image of the unit circle by the conformal transformation, if $w_{a}(z_{a})$ were analytic at $z_{a,1}$, then $w_{b}(z_{b})$ would be analytic at $z_{b,1}$, because the composition of two analytic functions is also an analytic function. Therefore, if $w_{b}(z_{b})$ is singular at $z_{b,1}$, then $w_{a}(z_{a})$ must be singular at $z_{a,1}$. This establishes Lemma 1.

Lemma 2   : If the singularity of $w_{b}(z_{b})$ at $z_{b,1}$ is a soft one, then the singularity of $w_{a}(z_{a})$ at the corresponding singular point $z_{a,1}$ under the conformal transformation is also a soft singularity.

Since according to the definition in Equations (16) we have that $w_{a}(z_{a})=w_{b}(z_{b})$ and since the $z_{b}\to z_{b,1}$ limit on the $z_{b}$ plane corresponds, through the continuous conformal transformation, to the $z_{a}\to z_{a,1}$ limit on the $z_{a}$ plane, it follows that if the limit

$$\lim_{z_{b}\to z_{b,1}} w_{b}(z_{b})$$ (17)

exists, then so does the limit

$$\lim_{z_{a}\to z_{a,1}} w_{a}(z_{a}).$$ (18)

Therefore, if the singularity of $w_{b}(z_{b})$ at $z_{b,1}$ is a soft one, which means that the first limit exists, then the singularity of $w_{a}(z_{a})$ at $z_{a,1}$ is also a soft singularity, since in this case the second limit also exists. This establishes Lemma 2.

Lemma 3   : If the singularity of $w_{b}(z_{b})$ at $z_{b,1}$ is a hard one, then the singularity of $w_{a}(z_{a})$ at the corresponding singular point $z_{a,1}$ under the conformal transformation is also a hard singularity.

By an argument similar to the one used in Lemma 2, since according to the definition in Equations (16) we have that $w_{a}(z_{a})=w_{b}(z_{b})$, and since the $z_{b}\to z_{b,1}$ and $z_{a}\to z_{a,1}$ limits correspond to one another, if $w_{b}(z_{b})$ is not well defined at $z_{b,1}$, which means that the limit

$$\lim_{z_{b}\to z_{b,1}} w_{b}(z_{b})$$ (19)

does not exist, then the limit

$$\lim_{z_{a}\to z_{a,1}} w_{a}(z_{a})$$ (20)

also does not exist, and therefore $w_{a}(z_{a})$ cannot be well defined at the corresponding point $z_{a,1}$. Therefore, if the singularity of $w_{b}(z_{b})$ at $z_{b,1}$ is a hard one, then the singularity of $w_{a}(z_{a})$ at $z_{a,1}$ must also be a hard singularity. This establishes Lemma 3.

Lemma 4   : If the singularity of $w_{b}(z_{b})$ at $z_{b,1}$ is an integrable one, then the singularity of $w_{a}(z_{a})$ at the corresponding singular point $z_{a,1}$ under the conformal transformation is also an integrable singularity.

If the singularity of $w_{b}(z_{b})$ at $z_{b,1}$ is an integrable one, then the integral of $w_{b}(z_{b})$ along any open integration contour $D_{b}$ going from a point $z_{b,0}$ internal to $C_{b}$ to the singular point $z_{b,1}$,

$$\int_{D_{b}}dz_{b}\, w_{b}(z_{b}),$$ (21)

exists and is a finite complex number. We consider now the corresponding integral of $w_{a}(z_{a})$ along an arbitrary open integration contour $D_{a}$ going from an internal point $z_{a,0}$ within the open unit disk to the singular point $z_{a,1}$ on the unit circle, and make a transformation of the integration variable from $z_{a}$ to $z_{b}$,

$$\int_{D_{a}}dz_{a}\, w_{a}(z_{a}) = \int_{D_{b}}dz_{b}\, \left( \frac{dz_{a}}{dz_{b}} \right) w_{b}(z_{b})$$

$$= \int_{D_{b}}dz_{b}\, \left[ \frac{d\gamma^{(-1)}(z_{b})}{dz_{b}} \right] w_{b}(z_{b}),$$ (22)

where we used the relations shown in Equations (14) and (16), where $D_{b}$ is an open integration contour going from an internal point $z_{b,0}$ to the singular point $z_{b,1}$, and where $D_{b}$, $z_{b,0}$ and $z_{b,1}$ correspond respectively to $D_{a}$, $z_{a,0}$ and $z_{a,1}$, through the conformal transformation. Since the conformal transformation is an analytic function, the derivative within brackets is also an analytic function within and on $D_{b}$, and therefore is a limited function there. Since $w_{b}(z_{b})$ by hypothesis is an integrable function around the singular point $z_{b,1}$, and since the product of a limited function and an integrable function is also an integrable function, the integrand of this last integral is an integrable function, and therefore this last integral exists and is a finite complex number. We thus conclude that

$$\int_{D_{a}}dz_{a}\, w_{a}(z_{a})$$ (23)

exists and is a finite complex number. Therefore, $w_{a}(z_{a})$ is an integrable function around the singular point $z_{a,1}$, and therefore that singularity is also an integrable one. This establishes Lemma 4.

We must now consider the question of what is the set of curves $C_{b}$ for which the structure described above can be set up. Given the curve $C_{b}$, the only additional objects we need in order to do this is the conformal mapping $\gamma(z_{a})$ and its inverse $\gamma^{(-1)}(z_{b})$, between that curve and the unit circle $C_{a}$, as well as between the respective interiors. The existence of these transformation functions can be ensured as a consequence of the Riemann mapping theorem, and of the associated results relating to conformal mappings between regions of the complex plane [#!RMPQiu!#]. According to that theorem, a conformal transformation such as the one we just described exists between any bounded simply connected open set of the plane and the open unit disk, and can be extended to the respective boundaries so long as the curve $C_{b}$ is differentiable. There are therefore no additional limitations on the differentiable simple closed curves $C_{b}$ that may be considered here.