International Journal of Analysis and Applications Volume 19, Number 5 (2021), 674-694 URL: https://doi.org/10.28924/2291-8639 DOI: 10.28924/2291-8639-19-2021-674 APPLICATIONS ON THE BOUNDARY BEHAVIOUR OF THE DERIVATIVE OF CONFORMAL MAPPING SHATHA S. ALHILY1, VISHNU NARAYAN MISHRA2,∗ 1Department of Mathematics, College of Science, Mustansiriyah University, Baghdad, Iraq shathamaths@uomustansiriyah.edu.iq, shathamaths@yahoo.co.uk 2Department of Mathematics, Indira Gandhi National Tribal University, Lalpur, Amarkantak, Anuppur 484 887, India ∗Corresponding author: vishnunarayanmishra@gmail.com, vnm@igntu.ac.in Abstract. 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. 1. Introduction Let φ be a conformal map from D ⊂ C onto a simply connected domain Ω, with its inverse ψ = φ−1 : Ω −→ D. Brennan’s conjecture states that, for all such φ, (1.1) ∫ ∫ D |φ′|2−pdxdy = ∫ ∫ Ω |ψ′|pdxdy < ∞, for 4 3 < p < 4. In connection with this conjecture which is associated with estimating the integral means of derivatives of univalent functions, which is so related with the behaviour of the boundary derivative of conformal maps from polygon domain onto unit disk and the behaviour of the boundary of inverse maps Received March 7th, 2020; accepted August 3rd, 2020; published August 2nd, 2021. 2010 Mathematics Subject Classification. 30B30. Key words and phrases. conformal mapping; derivative of conformal maps; infinite sector; Polygon domain. ©2021 Authors retain the copyrights of their papers, and all open access articles are distributed under the terms of the Creative Commons Attribution License. 674 https://doi.org/10.28924/2291-8639 https://doi.org/10.28924/2291-8639-19-2021-674 Int. J. Anal. Appl. 19 (5) (2021) 675 An existence and finiteness the integrability of the derivative of conformal maps through constructing some interesting examples when engages the infinite sector W. 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 v1,v2, ...,vn and n edges v1v2,v2v3, ...,vn−1vn,vnv1 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(z0) = 0 and φ ′(z0) > 0 ( where z0 is some point in P), and observe that φ maps γ onto an arc γ̂ of the unit circle |w| = 1 with ψ(w) = φ−1(z) : D −→ P. φ(z) = A + C ∫ z n−1∏ k=1 (ζ −zk)αk−1dζ 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. Int. J. Anal. Appl. 19 (5) (2021) 676 This technique could help to study the behavior of conformal mapping by estimating some quantities which belong to interior upper half-plane, which in turn will help to analyze the boundary behaviour of the derivative of this map. 2. Main Results In the light of reported above, we discussed the theoretical underpinnings of the behaviour of the boundary derivative of conformal maps through of some main results obtained that have been yielded useful information in this regards. (cf. [1]) Theorem 2.1. The derivative of the conformal mapping defined on Rectangular domain to the Unit disk is bounded but the derivative of the inverse maps is unbounded. H+ -1 -a a 1 rectangle unit disk D ζ = snz z = arcsn(ζ,k) φ = sn z− i sn z+ i ψ = sn z+ i 2i cn zd n z ζ = i w + 1 w− 1 w = ζ− iζ+ i Figure 1. Conformal mappings from rectangular domain onto unit disk Proof. Map the interior of the rectangular domain with vertices at points z1 = 1, z2 = −1, z3 = −a, and z4 = a with a > 1 into upper half plane needs to define Schwarz- Christoffel transformtion which maps H+ Int. J. Anal. Appl. 19 (5) (2021) 677 into rectangle as follows: z = A + B ∫ ζ ζo ds (s− 1) 1 2 (s + 1) 1 2 (s−a) 1 2 (s + a) 1 2 z = A + B ∫ ζ ζo ds (s2 − 1) 1 2 (s2 −a2) 1 2 Suppose A = 0 and B = 1 for convenience. z = ∫ ζ ζo ds√ (s2 − 1)(s2 −a2) (2.1) Suppose ζo = 0 and let a = 1 k , then the integral (2.10) is transformed to z = ∫ ζ ζo ds√ (s2 − 1)(s2 − ( 1 k )2) = k ∫ ζ 0 ds√ (s2 − 1)(k2s2 − 1) z = k ∫ ζ 0 ds√ (1 −s2)(1 −k2s2) (2.2) 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 z = sn−1(ζ,k).(2.3) The inverse mapping of integral (2.2) is known as the Jacobi elliptic function denoted by ζ = sn(z,k). ⇒ φ = ζ − i ζ + i = snz − i snz + i : Rectangular domain −→ D φ′ = (snz + i)(cnz dnz) − (snz − i)( dnz) (snz + i)2 = 2i cnz dnz (snz + i)2 , such that, dnz = √ 1 −k2sn2z , cnz = √ 1 −sn2z. |φ′| = ∣∣∣∣2i cnz dnz(snz + i)2 ∣∣∣∣ = 2|cnz dnz||snz + i|2 snz ∈ H+ ⇒ snz + i ∈ H+ that is, |snz + i| ≥ 1 ⇒ 1 |snz + i| ≤ 1 |cnz dnz| = | √ 1 −k2sn2z| | √ 1 −sn2z| |1 −k2sn2z| ≤ 1 + |snz|2 |1 −sn2z| ≤ 1 + |k snz|2 ⇒ |φ′| = 2|cnz dnz| |snz + i|2 ≤ 2(1 + |snz|2)(1 + |k snz|2) ⇒ |φ′| is bounded. It remains to show that the inverse of the derivative of such function φ(z) is unbounded as follows: Int. J. Anal. Appl. 19 (5) (2021) 678 ψ = φ−1 : D −→ rectangular domain ψ′ = 1 φ′ = (snz + i)2 2i cnz dnz ⇒ ψ = (snz + i)2 2i cnz dnz |ψ′| = |snz + i|2 2|cnz dnz| snz ∈ H+ ⇒ (snz + i) ∈ H+ ⇒ |snz + i|2 ≥ 1. |ψ′| = |snz + i|2 2|cnz dnz| ≥ 1 2|cnz dnz| ⇒ |ψ′| is unbounded as |z| −→ 0. � Theorem 2.2. The derivative of the conformal mapping defined on Triangular domain M to the unit disk is bounded but the derivative of the inverse maps is unbounded. H+ z-plane w1 = 0 b w2 w3 = b w-plane ζ-plane D φ1 φ−11 h( w )φ 2 Figure 2. Conformal mappings from triangular domain onto unit disk Int. J. Anal. Appl. 19 (5) (2021) 679 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 M by Schwarz-Christoffel transformation as follows: Let φ1 = A ∫ z zo (s−x1)−k1 (s−x2)−k2ds + B. such that −ki = αiπ − 1 ; ∀i = 1, 2 to be Schwars-Christoffel transformation that maps H + into the interior of the equilateral triangle M such that αi = π 3 ; ∀i = 1, 2, 3. Now, by assisstance that zo = 1, A = 1 and B = 0, we obtain Schwars-Christoffel transformation defined as follows: φ1 = ∫ z 1 (s + 1) −2 3 (s− 1) −2 3 ds.(2.4) Which maps x1 = −1,x2 = 1 and x3 = ∞ into M w1w2w3 as follows: (i) In case z = 1 ⇒ φ1(1) = 0 ; that is , w1 = 0 in H+. (ii) In case z = −1, we have φ1(−1) = ∫ −1 1 (s + 1) −2 3 (s− 1) −2 3 ds = w2.(2.5) (iii) In case z = ±∞, we have lim z→±∞ φ1 = ∫ ±∞ 1 (s + 1) −2 3 (s− 1) −2 3 ds = w3.(2.6) 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 s− 1 = |s− 1| eiθ1 s + 1 = |s + 1| eiθ2. The argument (θ1 + θ2) 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 w2 = φ1(−1) = − ∫ 1 −1 (x + 1) −2 3 (1 −x) −2 3 (−e− 2πi 3 )ds. w2 = φ1(−1) = e πi 3 ∫ 1 0 2dx (1 −x2) −2 3 .(2.7) Int. J. Anal. Appl. 19 (5) (2021) 680 By letting x = √ t in equation (2.7), we obtain Beta function B( 1 2 , 1 3 ) and w2 = φ1(−1) = e πi 3 B( 1 2 , 1 3 ). ⇒ w2 = b e πi 3 . where b is the value of B( 1 2 , 1 3 ) Now, the vertex w3 lies on the positive u- axis. So, w3 must be represented by the boundary integral 1 to ∞ as follows, w3 = φ1(∞) = ∫ ∞ 1 (x + 1) −2 3 (1 −x) −2 3 dx w3 = φ1(∞) = ∫ ∞ 1 dx (x2 − 1) −2 3 .(2.8) But w3 is also represented by integral (2.4) when z = −∞ along the negative real axis. So, w3 = φ1(−∞) = ∫ −∞ 1 (x + 1) −2 3 (1 −x) −2 3 dx = ∫ −1 1 (x + 1) −2 3 (1 −x) −2 3 e −2πi 3 dx + ∫ −∞ −1 (x + 1) −2 3 (1 −x) −2 3 e −4πi 3 dx. w3 = w1 + e −πi 3 ∫ ∞ 1 dx (x2 − 1) 2 3 ⇒ w3 = w1 + e −πi 3 w3.(2.9) Solving (2.9) for w3 we obtain: ⇒ w3 −e −πi 3 w3 = w1. w3 (1 −e −πi 3 ) = b e −πi 3 w3 = b ; since (1 −e −πi 3 ) = e −πi 3 . In the end, we found the conformal mapping that maps upper half- plane H+ onto triangular domain M. It is known that φ2 = z−i z+i maps upper half- plane H+ onto unit disk D. Hence, we have φ1 = ∫ z 1 (s2 − 1) −2 3 ds : upper half plane H+ −→ triangular domain M and φ2 = z − i z + i : upper half- plane H+ → unit diskD. Let h(w) : triangular domain M−→ unit disk D. Int. J. Anal. Appl. 19 (5) (2021) 681 defined as follows: h(w) = (φ2 ◦φ−11 )(w) = φ2(φ −1 1 )(w) h′(w) = φ′2(φ −1 1 )(φ ′−1 1 ) = ( φ−11 − i φ−11 + i ) ′ ( 1 φ′1 ) = (φ−11 + i)(φ ′−1 1 ) − (φ −1 1 − i)(φ ′−1 1 ) (φ−11 + i) 2 1 φ′1 . ⇒ h′(w) = 2i(z2 − 1) 2 3 (φ−11 + i) 2 (z2 − 1) 2 3 = 2i(z2 − 1) 4 3 (φ−11 + i) 2 . |h′(w)| = | 2i(z2 − 1) 4 3 (φ−11 + i) 2 | = 2i|z2 − 1| 4 3 |φ−11 + i|2 , where (φ−11 ) ′ = 1 φ′1 = (z2 − 1) 2 3 and φ−11 ∈ H +, so that (φ−11 + i) ∈ H +. This implies to |φ−11 + i| 2 ≥ 1 ⇒ 1 |φ−11 + i|2 ≤ 1 ⇒ |h′(w)| ≤ 2(|z|2 + 1) 4 3 . |h′(w)| is bounded. What remains is to prove that the inverse of the derivative of h(w) is unbounded. Note that; φ−12 = −i ζ+1 ζ−1 is the Möbius transformation, it maps the unit disk D to upper half-plane H +, thus we have (φ−12 ) ′ = (ζ − 1)(−i) − (−iζ − i) (ζ − 1)2 = 2i (ζ − 1)2 . We use Schwarz-Christoffel transformation for mapping H+ into triangular domain M. φ1 = ∫ z 1 (s2 − 1) −2 3 ds. Int. J. Anal. Appl. 19 (5) (2021) 682 Therefore, h−1(ζ) = (φ1 ◦φ−12 )(ζ) : D −→M . (h−1(ζ)) ′ = φ′1(φ −1 2 (ζ)) (φ −1 2 (ζ)) ′ = [(φ−12 (ζ)) 2 − 1] −2 3 2i (ζ − 1)2 . = 2i [ −( ζ+1 ζ−1 ) 2 − 1 ]−2 3 (ζ − 1)2 = 2i [ −2ζ2−2 (ζ−1)2 ]−2 3 (ζ − 1)2 . = 2i (ζ − 1)2 [ −2ζ2−2 (ζ−1)2 ]−2 3 = 2i (ζ − 1)2 (−2ζ2 − 2) 2 3 (ζ − 1) −4 3 = 2i (ζ − 1) 2 3 (−2ζ2 − 2) 2 3 . |(h−1(ζ)) ′ | = ∣∣∣∣ 2i (ζ − 1) 2 3 (−2ζ2 − 2) 2 3 ∣∣∣∣ . |(h−1(ζ)) ′ | = ∣∣∣∣ 2 (ζ − 1) 2 3 (2ζ2 + 2) 2 3 ∣∣∣∣ −→∞ as ζ −→ 1. � Theorem 2.3. The derivative of the conformal mapping defined on crescent domain to unit disk is bounded but the derivative of the inverse maps is unbounded. Proof. Compute the conformal mapping of a crescent domain onto unit disk by setting a sequence of functions as follows: Let φ1 = 1 z maps C1 → L1 and C2 → L2. φ2 = i z rotates the stripe in the left plane onto stripe in the lower half plane. φ3 = 4πi z extends the stripe in the lower half plane H− between −π,−2π φ4 = e 4iπ z maps the stripe in the lower half plane H−into H+. φ5 = e 4iπ z − i e 4iπ z + i maps the H+ onto unit disk D. Int. J. Anal. Appl. 19 (5) (2021) 683 −2 c2 −4 crescent domain c1 L1 −1 4 L2 −1 2 −1 4 −1 2 −π −4π H+ unit disk D φ1 φ2 φ3 φ4 φ5 Figure 3. Conformal mappings from crescent domain onto unit disk So; φ(z) = e 4iπ z − i e 4iπ z + i : crescent domain −→ unit disk D. Using short calculation we obtain φ′(z) = (e 4iπ z + i)(−4iπ z2 e 4iπ z ) − (e 4iπ z − i)(−4iπ z2 e 4iπ z ) (e 4iπ z + i)2 . ⇒ φ′(z) = 8π z2 e 4iπ z (e 4iπ z + i)2 = 8πe 4iπ z z2(e 4iπ z + i)2 . |φ′(z)| = 8π|e 4iπ z | |z|2|e 4iπ z + i|2 . Int. J. Anal. Appl. 19 (5) (2021) 684 Now; |e 4iπ z | = 1 ⇒ e 4iπ z ∈ H+ ⇒ (e 4iπ z + i) ∈ H+. ⇒ |e 4iπ z + i| ≥ 1 ⇒ |e 4iπ z + i|2 ≥ 1 ⇒ 1 |e 4iπ z + i| 2 ≤ 1. ⇒ |φ′(z)| = 8π|e 4iπ z | |z|2|e 4iπ z + i|2 . ≤ 8π |z|2 Hence, |φ′(z)| is bounded. We show that the inverse of the derivative of such function φ(z) is unbounded as follows: φ−1(w) = ψ(w) : D −→ crescent domain. ψ = φ−1 = 1 φ = e 4πi z + i e 4πi z − i . ⇒ ψ′ = 1 φ′ ⇒ ψ′ = z2(e 4iπ z + i) 8πe 4iπ z ⇒ |ψ′| = |z|2|e 4iπ z + i| 8π|e 4iπ z | . It is known, e 4iπ z ∈ H+ ⇒ (e 4iπ z + i) ∈ H+. Hence |e 4iπ z + i| ≥ 1 & |e 4iπ z | = 1. ⇒ |ψ′| = |z|2|e 4iπ z + i| 8π|e 4iπ z | ≥ |z|2 8π . 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: Int. J. Anal. Appl. 19 (5) (2021) 685 α β 0 H+ 0-1 1 unit disk D -1 1 i −i φ1 φ2 φ3 Figure 4. Conformal mappings from lens domain onto unit disk Let φ1 = z −α z −β maps Lens-shaped domain to the first quarter plane φ2 = z 2 maps the first quarter plane to the upper half- plane. φ3 = z2 − i z2 + i maps upper half plane H+ to the unit disk D. So, φ(z) = (z−α z−β ) 2 − i (z−α z−β ) 2 + i : Lens-shaped domain −→ unit disk D. ⇒ φ(z) = (z −α)2 − i(z −β)2 (z −α)2 + i(z −β)2 . Therefore when z = α ⇒ φ(z) = −1 in D z = β ⇒ φ(z) = 1 in D. Int. J. Anal. Appl. 19 (5) (2021) 686 When (z −α)2 = −(z −β)2 ⇒ φ(z) = −(z −β)2 − i(z −β)2 −(z −β)2 + i(z −β)2 = i in D. In the end, if (z −α)2 = (z −β)2 ⇒ φ(z) = (z −β)2 − i(z −β)2 (z −β)2 + i(z −β)2 = −i in D. ⇒ φ′(z) = 4i(z −β)2(z −α) − 4i(z −β)(z −α)2 [(z −α)2 + i(z −β)2]2 ⇒ |φ′(z)| = 4|α−β||z −β||z −α| (z −α)4 + (z −β)4 . where, 4|α−β| = c; c is a constant; (α < β). Also, |z −β| ≤ |z| + |β| = M1 & |z −α| ≤ |z| + |α| = M2 ⇒ |φ′(z)| ≤ M (z −α)4 + (z −β)4 ⇒ |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: ψ′ = 1 φ′ = [(z −α)2 + i(z −β)2]2 4i(z −β)2(z −α) − 4i(z −β)(z −α)2 ⇒ |ψ′| = |(z −α)2 + i(z −β)2|2 4|(z −β)(z −α)(α−β)| ⇒ |ψ′| = (z −α)4 + (z −β)4 4|z −β||z −α||α−β| . 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. Int. J. Anal. Appl. 19 (5) (2021) 687 Let φ(z) = (φ2 ◦φ1)(z) = z2 − i z2 + i : W → D so that φ1(z) = z2. If maps the infinite sector onto upper half plane H+ and φ2(w) = w−iw+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 > 2 3 . !"#$%&#'(%)*+,-."/% 0%1234% 56% 54% 7+,-."/% Figure 5. Infinite sector W for the angle α = π 2 Proof. Given φ(z) = z2 − i z2 + i : W → D ⇒ φ′(z) = (z2 + i)(2z) − (z2 − i)(2z) (z2 + i)2 = i4z (z2 + i)2 ⇒ |φ′(z)| = 4|z| |z2 + i|2 . Now, W-plane is an infinite sector. that is; r = |z|→ 0 −∞. (i) so, r = |z|→∞ (i.e; |z| is large). We know that |z2 + i| ≥ |z|2 − 1 ≥ 1 2 |z|2. ⇒ 1 |z2 + i| ≤ 2 |z|2 . ⇒ 1 |z2 + i|2 ≤ 4 |z|4 , Int. J. Anal. Appl. 19 (5) (2021) 688 refer to the behaviour of |z2 + i| at ∞ with respect to the region. ⇒ |φ′(z)| = 4|z| |z2 + i|2 ≤ 16|z| |z|4 = 16|z|−3. (ii) and r = |z| ∼ 0 ⇒ |φ′(z)| = 4|z| |z2 + i|2 ⇒ |φ′(z)| = 4|z| |i|2 = 4|z|. ⇒ |φ′(z)| ≤   16|z| −3 ; |z| is large 4|z| ; |z| ∼ 0 ∫ π 2 0 ∫ ∞ 0 |φ′|p dxdy ≤ 4 ∫ π 2 0 ∫ 1 0 |z|p r drdθ + 16 ∫ π 2 0 ∫ ∞ 1 |z|−3p r drdθ. ∫ π 2 0 ∫ ∞ 0 |φ′|p dxdy = 2π [∫ 1 0 rp+1 dr ] + 8π [∫ ∞ 1 r−3p+1 dr ] . ⇒ ∫ π 2 0 ∫ ∞ 0 |φ′|p dxdy ≤ 2π [ rp+2 p + 2 ∣∣∣∣1 0 + 2π [ r−3p+2 −3p + 2 ∣∣∣∣∞ 1 (2.10) There are two definite integrals on the right-hand side of inequality (2.10). The first one is clearly finite, and the second one is: −3p + 2 < 0 ⇒ −3p < −2 ⇒ p > 2 3 . � Theorem 2.6. Let φ(z) be a conformal mapping defined on infinite sector W onto unit disk D. Let φ(z) = (φ2 ◦φ1)(z) = z4 − i z4 + i : W → D such that φ1(z) = z 4 maps the infinite sector onto upper half plane H+ and φ2(w) = w−iw+i maps the upper half plane H+ onto unit disk D (see Figure 6). then the integrability of the derivative of conformal mapping is: ∫ ∫ W |φ′(z)|p dxdy < ∞ ; for each p > 2 5 . Int. J. Anal. Appl. 19 (5) (2021) 689 !"##!$# %&'()*+# ,-&'()*+# .*/0#1/23# 45678# Figure 6. Infinite sector W for the angle α = π 4 Proof. Given φ(z) = z4 − i z4 + i : W → D ⇒ φ′(z) = (z4 + i)(4z3) − (z4 − i)(4z3) (z4 + i)2 = i8z3 (z4 + i)2 ⇒ |φ′(z)| = 8|z|3 |z4 + i|2 . Now, W-plane is an infinite sector. that is; r = |z|→ 0 −∞. • so,when r = |z|→∞ (that is; |z| be large). We know that |z4 + i| ≥ |z|4 − 1 ≥ 1 2 |z|4. ⇒ 1 |z4 + i| ≤ 2 |z|4 . ⇒ 1 |z4 + i|2 ≤ 4 |z|8 . This is referring to the behaviour of |z4 + i| at ∞ with respect to the region. ⇒ |φ′(z)| = 8|z|3 |z4 + i|2 ≤ 8|z|3 |z|8 = 32|z|−5. • and when r = |z| ∼ 0 ⇒ |φ′(z)| = 8|z|3 |z4 + i|2 ⇒ |φ′(z)| = 8|z|3 |i|2 = 8|z|3. ⇒ |φ′(z)| ≤   8|z| −5 ; |z| is large 8|z|3 ; |z| ∼ 0 Int. J. Anal. Appl. 19 (5) (2021) 690∫ π 4 0 ∫ ∞ 0 |φ′|p dxdy ≤ 8 ∫ π 4 0 ∫ 1 0 r3pr drdθ + 8 ∫ π 4 0 ∫ ∞ 1 r−5pr drdθ = 2π [∫ 1 0 r3p+1 dr ] + 2π [∫ ∞ 1 r−5p+1 dr ] ⇒ ∫ π 4 0 ∫ ∞ 0 |φ′|p dxdy ≤ 2π [ r3p+2 3p + 2 |10 + 2π [ r−5p+2 −5p + 2 |∞1(2.11) The first term on the right -hand- side of (2.11) is finite, and the second one is: −5p + 2 < 0 ⇒ −5p < −2 ⇒ p > 2 5 . � Theorem 2.7. Let φ(z) be a conformal mapping defined on infinite sector W onto unit disk D. Let φ(z) = (φ2 ◦φ1)(z) = zn − i zn + i : W → D such that φ1(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 φ2(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:∫ ∫ W |φ′(z)|p dxdy < ∞ ; for each p > 2 3π α + 1 . !"##!$# %&'()*+# ,-&'()*+# .*/0#1/23# 4567*# Figure 7. Infinite sector W for the angle α = π n Proof. Let α = π n be the angle of the infinite sector W which is mapped by φ1 = z n onto upper half plane H+. One can write φ1 = z π α . We define the power function φ1 = z π α to be the multivalued function z π α = e π α log z ; z 6= 0. Int. J. Anal. Appl. 19 (5) (2021) 691 ⇒ z π α = e π α log |z|+iargz = r π α ei π α θ e±i 2π2 α k Various values of z π α are obtained from the principal value e π α log z by multiplying by the integral power (ei 2π2 α ) k of ei 2π2 α . Let α = π n is a number for some integer n, then the integral powers ei 2π2 α k of ei 2π2 α are exactly the nth roots of unity, and the values of z π α are the n nth roots of z. ⇒ φ(z) = z π α − i z π α + i = r π α ei π α − i r π α ei π α + i φ(z) = r π α cos π α θ + ir π α sin π α θ − i r π α cos π α θ + ir π α sin π α θ + i . (2.12) Simplify last equation (2.12) we get: φ(z) = r 2π α − 2ir π α cos π α θ − 1 r 2π α + 2r π α sin π α θ + 1 = r 2π α − 1 r 2π α + 2r π α sin π α θ + 1 + i −2r π α cos π α θ r 2π α + 2r π α sin π α θ + 1 . When θ = 0 or θ = 2π, it implies that: φ(z) = r 2π α − 1 r 2π α + 1 + i −2r π α r 2π α + 1 = r 2π α − 1 − 2ir π α r 2π α + 1 The derivative of φ(z) can be calculated: φ′(z) = (r 2π α + 1)[ 2π α r 2π α −1 − 2iπ α r π α −1] − (r 2π α − 1 − 2ir π α )[ 2π α r 2π α −1] (r 2π α + 1)2 = 2iπ α r 3π α −1 + 4π α r 4π α −1 − 2iπ α r π α −1 (r 2π α + 1)2 = 4π α r 4π α −1 + i2π α (r 3π α −1 −r π α −1) (r 2π α + 1)2 |φ′(z)| = ∣∣∣4πα r 4πα −1 + i2πα (r 3πα −1 −rπα−1)∣∣∣ (r 2π α + 1)2 |φ′(z)| = √ 16π2 α2 r 8π α −2 + 4π 2 α2 (r 3π α −1 −r π α −1)2 (r 2π α + 1)2 Int. J. Anal. Appl. 19 (5) (2021) 692 Again, W-plane is an infinite sector. that is; |z|→ 0 −∞. (i) In case r = |z|→∞ (that is; |z| be large). |φ′(z)| = √ 16π2 α2 |z| 8π α −2 + 4π 2 α2 (|z| 3π α −1 −|z| π α −1 )2 (|z| 2π α + 1)2 . We know that, |z| 2π α + 1 ≥ |z 2π α + 1| ≥ |z| 2π α − 1 ≥ 1 2 |z| 2π α ⇒ |z| 2π α + 1 ≥ 1 2 |z| 2π α ⇒ 1 (|z| 2π α + 1)2 ≤ 4 |z| 4π α . Hence, |φ′(z)| = √ 16π2 α2 |z| 8π α −2 + 4π 2 α2 (|z| 3π α −1 −|z| π α −1 )2 (|z| 2π α + 1)2 ≤ 4 √ 16π2 α2 |z| 8π α −2 + 4π 2 α2 (|z| 3π α −1 −|z| π α −1 )2 |z| 4π α . |φ′(z)| ≤ 4 √ 16π2 α2 r 8π α −2 + 4π 2 α2 (r 3π α −1 −r π α −1)2 r 4π α = 4 √ 16π2 α2 r 8π α −2 + 4π 2 α2 r 2π α −2(r 2π α − 1)2 r 4π α = 4 √ 4π2 α2 r 2π α −2 [ 4r 6π α + (r 2π α − 1)2 ] r 4π α |φ′(z)| ≤ 4 √ 4π2 α2 r 2π α −2 [ 4r 6π α + (r 2π α − 1)2 ] r 4π α = 8π α r π α −1 √ 4 r 6π α + (r 2π α − 1)2 r 4π α = 8π α r −3π α −1 √ 4 r 6π α + (r 2π α − 1)2 (ii) In case r = |z| ∼ 0 |φ′(z)| = √ 16π2 α2 |z| 8π α −2 + 4π2 α2 (|z| 3π α −1 −|z| π α −1 )2. ⇒ |φ′(z)| = √ 16π2 α2 r 8π α −2 + 4π2 α2 (r 3π α −1 −r π α −1)2 = √ 16π2 α2 r 8π α −2 + 4π2 α2 r 2π α −2(r 2π α − 1)2 = √ 4π2 α2 r 2π α −2 [ 4r 6π α + (r 2π α − 1)2 ] = 2π α r π α −1 √ 4 r 6π α + (r 2π α − 1)2. Int. J. Anal. Appl. 19 (5) (2021) 693 In the end, ⇒ |φ′(z)| ≤   8π α r −3π α −1 √ 4 r 6π α + (r 2π α − 1)2 ; |z| is large 2π α r π α −1 √ 4 r 6π α + (r 2π α − 1)2. ; |z| ∼ 0 ∫ π n 0 ∫ ∞ 0 |φ′(z)|pdxdy ≤ ∫ π n 0 ∫ 1 0 ( 2π α )p r( π α −1)p [ 4 r 6π α + (r 2π α − 1)2 ]p 2 .rdrdθ + ∫ π n 0 ∫ ∞ 1 ( 8π α )p r( −3π α −1)p [ 4 r 6π α + (r 2π α − 1)2 ]p 2 rdrdθ. ∫ π n 0 ∫ ∞ 0 |φ′(z)|pdxdy ≤ ( 2π α )p( π n ) ∫ 1 0 r( π α −1)p+1dr ∫ 1 0 [ 4 r 6π α + (r 2π α − 1)2 ]p 2 .rdr +( 8π α )p( π n ) ∫ ∞ 1 r( −3π α −1)p+1dr ∫ ∞ 1 [ 4 r 6π α + (r 2π α − 1)2 ]p 2 dr. such that, we have four terms . The first, second and fourth terms on the right -hand-side of (2.4) becomes finite only when ( −3π α − 1)p + 2 ≤ 0 ⇒ ( −3π α − 1)p ≤−2 ⇒ −( 3π α + 1)p ≤−2 ⇒ ( 3π α + 1)p ≥ 2 ⇒ p > 2 3π α + 1 . � Acknowledgment: The author would like to express a deep thanks and gratitude to Department of Math- ematics , College of Science, Mustansiriyah University for deep supporting to appear this research paper as it is now. Conflicts of Interest: The author(s) declare that there are no conflicts of interest regarding the publication of this paper. References [1] S. S. Alhily, Higher Integrability of the Gradient of Conforaml Maps, PhD thesis, University of Sussex, 2013. [2] S. S. Alhily and Deepmala, Boundary Behaviour of Holomorphic Functions on the Cardioid Domain with some Applications, Bol. Soc. Paran. Mat. 38(7) (2020), 203-218. [3] P. L. Duren, Univalent Functions, third ed., Springer-Verlag, New York, 1983. [4] P. H. Difrancesco, P. Mathieu, D. Sénéchal, Conformal Field Theory, first ed., Springer-Verlag Berlin Heidelberg, New York, 1997. Int. J. Anal. Appl. 19 (5) (2021) 694 [5] T. A. Driscoll, L. N. Trefethen, Schwarz-Christoffel Mapping. Cambridge Monographs on Applied and Computational Mathematics, vol. 8. Cambridge University Press, Cambridge, 2002. [6] S. G. Krantz, Complex Analysis: The Geometric Viewpoint, vol.23, Mathematical Association of America, Washington, D.C., 1990. [7] V.N. Mishra and G. Tomar, Exiistance of wandering and periodic domain in given angular region, Math. Slovaca, 70(4) (2020), 839-848. [8] Ch. Pommerenke, Univalent functions, first ed., Vandenhoeck and Ruprecht, Göttingen, Germany, 1975. [9] Ch. Pommerenke, Boundary Behaviour of Conformal Mappings, second ed., Springer-Verlag Berlin Heidelberg, 1992. [10] E. M. Stein, R. Shakarchi, Complex Analysis. Princeton University Press, Princeton, 2003. [11] H. A. Saleh, S. S. Alhily, Growth and Bounded Solution of Second-Order of Complex Differential Equations Through of Coefficient Function, IOP Conf. Ser.: Mater. Sci. Eng. 871 (2020) 012057. [12] G. Tomar and V.N. Mishra, Maximum term of transcendental entire function and Spider’s web, Math. Slovaca, 70(4) (2020), 81-86. 1. Introduction 2. Main Results References