APPLICATIONS ON THE BOUNDARY BEHAVIOUR OF THE DERIVATIVE OF CONFORMAL MAPPING

. The objective of this research paper is to describe the behaviour of the boundary derivative of conformal maps from polygon domain onto unit disk through construct some an interesting cases, and its inverse maps. Moreover, we study the existence and ﬁniteness the integrability of the derivative of conformal maps over an inﬁnite sector W


Introduction
Let φ be a conformal map from D ⊂ C onto a simply connected domain Ω, with its inverse Brennan's conjecture states that, for all such φ, We started with basic definition of the polygonal domain and we present some typical examples to examine the boundary behaviour of the derivatives of conformal mapping of polygon to unit disk, as well as the boundary behaviour of the derivatives of the inverse maps.
In the following some preliminary on the boundary derivative of conformal maps, with basic definition of the polygonal domain that serves as motivation to the main results of this work to give a full support, the reader is referred to references [2, 3, 7-9, 11, 12].
Definition 1.1 (Polygon). A polygon P is usually defined as a collection of n vertices v 1 , v 2 , ..., v n and n edges v 1 v 2 , v 2 v 3 , ..., v n−1 v n , v n v 1 such that no pair of nonconsecutive edges share a point. We deviate from the usual practice by defining a polygon as the closed finite connected region of the plane bounded by these vertices and edges. The collection of vertices and edges will be referred to as the boundary of P , denoted by ∂P, a polygon of n vertices will sometimes be called an n-gon.
Riemann mapping theorem [6]. guarantees the existence of a conformal map φ from polygonal domain P ⊂ C conformally onto the unit disk D (|w| < 1), which can be extended continuously to the boundary ( cf. Carathéodory's theorem [9]). Worthy to mention that it is not yet possible to write down a simple formula for the conformal map from one region to another. Hence, in case of a map from the upper half-plane or unit disk D to a polygon, then Schwarz-Christoffel formula [5] allows to compute the conformal map φ defined as follows: Consider φ : P −→ D, be a conformal mapping, where ∂P be a circular arc or straight line segment γ, normalized by the conditions φ(z 0 ) = 0 and φ (z 0 ) > 0 ( where z 0 is some point in P ), and observe that φ maps γ onto an arcγ of the unit circle |w| = 1 with ψ(w) = φ −1 (z) : D −→ P.
where A and C are suitably chosen constants.
So,we deliberately construct φ as the composition of one Schwarz-Christoffel map from P into upper half-plane (by applying Schwarz-Christoffel transformation) and another map of the upper half-plane to unit disk, in the examples (2.1), (2.2) where φ maps the P to unit disk.
The term "polygon" is often modified by "simple" to distinguish it from polygons that cross themselves, Simple polygons are also called Jordan polygons, because such a polygon divides the plane into two regions. The boundary of a polygon is a "Jordan curve" : it separates the plane into two disjoint regions, the interior and the exterior of the polygon.
which belong to interior upper half-plane, which in turn will help to analyze the boundary behaviour of the derivative of this map.

Main Results
In the light of reported above, we discussed the theoretical underpinnings of the behaviour of the boundary The integral (2.2) is called an elliptic integral of first kind and k is a modulus of the elliptic integral with 0 < k < 1, denote by The inverse mapping of integral (2.2) is known as the Jacobi elliptic function denoted by ζ = sn(z, k).
It remains to show that the inverse of the derivative of such function φ(z) is unbounded as follows: H + z-plane Figure 2. Conformal mappings from triangular domain onto unit disk Proof. To construct conformal mapping defined on triangle domain to unit disk D. we have to define conformal mapping on triangular domain to upper half plane H + and then define another mapping from H + to unit disk D to achieve our aim.
To do so, First : we have to establish conformal mapping that maps upper half-plane H + onto triangular domain by Schwarz-Christoffel transformation as follows: such that −k i = αi π − 1 ; ∀i = 1, 2 to be Schwars-Christoffel transformation that maps H + into the interior of the equilateral triangle such that α i = π 3 ; ∀i = 1, 2, 3. Now, by assisstance that z o = 1, A = 1 and B = 0, we obtain Schwars-Christoffel transformation defined as follows: Which maps x 1 = −1, x 2 = 1 and x 3 = ∞ into w 1 w 2 w 3 as follows: To solve these integrals, let us consider first the equation (2.5) by choosing a path of the integration z = x along the real axis in the positive sense, that is The argument (θ 1 + θ 2 ) remains constant throughout integration from -1 to 1 since (s + 1) stays positive with zero argument, and (s − 1) has constant argument π. Therefore equation (2.5) yields 3 )ds.
where b is the value of B( 1 2 , 1 3 ) Now, the vertex w 3 lies on the positive u-axis. So, w 3 must be represented by the boundary integral 1 to ∞ as follows, But w 3 is also represented by integral (2.4) when z = −∞ along the negative real axis. So, dx.
Solving (2.9) for w 3 we obtain: In the end, we found the conformal mapping that maps upper half-plane H + onto triangular domain .
It is known that φ 2 = z−i z+i maps upper half-plane H + onto unit disk D. Hence, we have defined as follows: |h (w)| is bounded.
What remains is to prove that the inverse of the derivative of h(w) is unbounded.
Note that; φ −1 2 = −i ζ+1 ζ−1 is the Möbius transformation, it maps the unit disk D to upper half-plane H + , thus we have We use Schwarz-Christoffel transformation for mapping H + into triangular domain .    Using short calculation we obtain Hence, |φ (z)| is bounded. We show that the inverse of the derivative of such function φ(z) is unbounded as follows: In the end, we obtain |ψ | is unbounded.
Theorem 2.4. The derivative of the conformal mapping defined on Lens-shaped domain to the unit disk is bounded but the derivative of the inverse maps is unbounded.
Proof. Compute the conformal mapping a lens domain onto unit disk by setting a sequence of functions as follows: So, Therefore when In the end, if where, 4|α − β| = c; c is a constant; (α < β). Also, ⇒ |φ (z)| is a bounded for every z in Lens-shaped domain. Again, we can show that the inverse of the derivative of such function φ(z) is unbounded as follows: If z −→ α or z −→ β, then |ψ | −→ ∞, so |ψ | is unbounded.
The following examples show the integrability of the derivative of conformal maps on infinite sector W exists and is finite for some pth-power integrable function φ when α = π n is a number for some integer n.
Further details can be found in the books of Di Francesco [4] and of M. Stein [10].
Theorem 2.5. Let φ(z) be a conformal mapping defined on infinite sector W for the angle α = π 2 onto unit disk D.
If maps the infinite sector onto upper half plane H + and φ 2 (w) = w−i w+i maps the upper half-plane H + onto unit disk D (see Figure 5). Then the integrability of the derivative of conformal mapping φ, is as follows: W |φ (z)| p dxdy < ∞; for each p > Proof. Given Now, W-plane is an infinite sector. that is; r = |z| → 0 − ∞.
(i) so, r = |z| → ∞ (i.e; |z| is large). We know that refer to the behaviour of |z 2 + i| at ∞ with respect to the region.
(ii) and r = |z| ∼ 0 There are two definite integrals on the right-hand side of inequality (2.10).
The first one is clearly finite, and the second one is: Theorem 2.6. Let φ(z) be a conformal mapping defined on infinite sector W onto unit disk D. Let such that φ 1 (z) = z 4 maps the infinite sector onto upper half plane H + and φ 2 (w) = w−i w+i maps the upper half plane H + onto unit disk D (see Figure 6). then the integrability of the derivative of conformal mapping is: W |φ (z)| p dxdy < ∞ ; for each p > Proof. Given Now, W-plane is an infinite sector. that is; r = |z| → 0 − ∞.
• so,when r = |z| → ∞ (that is; |z| be large). We know that This is referring to the behaviour of |z 4 + i| at ∞ with respect to the region.
• and when r = |z| ∼ 0 The first term on the right -hand-side of (2.11) is finite, and the second one is: Theorem 2.7. Let φ(z) be a conformal mapping defined on infinite sector W onto unit disk D. Let such that φ 1 (z) = z n maps the infinite sector onto upper half-plane H + where α is of the form α = π n for some integer n and φ 2 (w) = w−i w+i maps the upper half-plane H + onto unit disk D (see Figure 7). Then the integrability of the derivative of conformal mapping is as follows: Proof. Let α = π n be the angle of the infinite sector W which is mapped by φ 1 = z n onto upper half plane H + .
We define the power function φ 1 = z π α to be the multivalued function Let α = π n is a number for some integer n, then the integral powers e i 2π 2 α k of e i 2π 2 α are exactly the nth roots of unity, and the values of z π α are the n nth roots of z.
We know that, Hence, such that, we have four terms . The first, second and fourth terms on the right -hand-side of (2.4) becomes finite only when