ANALYSIS OF THE FUNCTION AT INFINITY

Аннотация

This article describes methods for solving integrals using Function deduction with respect to apoint at infinity. This article can be agreat guide for independent learners Keywords:Complex area, deduction ,continuous, integral, curved line. Let the point at infinity z = be an isolated singular point of the function f (z). Let { } U( ) = z: z > R be a neighborhood of a point at infinity and let f (z) be analytic at U( ).Denote by C theclosed contour that liesentirely in U( ). Definition. The residue of the function f (z) with respect to the infinitely distant 1 point is the value of the integral f (z)dz, where the integration along the contour C 2pi C- occurs inthe negativedirection. TheLaurent series expansion in U( )for thefunction f (z) has theform c c f (z)=c + -1 + -2 + +c z +c z2 + 0 z z2 1 2 Since this series converges uniformly on the contour C , we can integrate it term by term. Noticingthat dz c zndz =0,если n 1; c =c 2pi n -1 z -1 C- C- weget after integration 1 f (z)dz =-c 2pi -1 C- thus,theresidue of afunction with respect to apoint at infinity resf ( ) = -c . -1 Comment. In the case of a removable singular point lying at a finite distance, the residueis always zero. This may not bethecase for apoint at infinity. page1110 INTERNATIONAL JOURNAL OF ARTIFICIAL INTELLIGENCE ISSN:2692-5206,ImpactFactor:12,23 AmericanAcademicpublishers,volume05,issue04,2025 Journal: https://www.academicpublishers.org/journals/index.php/ijai 1 For example, the function at infinity has a removable singularity, and the corresponding z residueis -1. Theorem. If f (z)is analytic at any point of the extended complex plane except for a finite number of singular points, then the sum of the residues with respect to all its singular points(including thepoint at infinity) isalways equal to zero. Proof.Let us describe a circle of finiteradius such that all singular points fall into this 1 circle. According tothe main residue theorem, the valueof f (z)dz isequal tothesum 2pi C ofresidues withrespect toall singular points lying inside C. 1 C on the other hand, the value f (z)dz is equal to the residue of the 2pi C- f (z) function with respect to the point at infinity. Therefore, the sum of all calculations is equal to 1 1 f (z)dz + f (z)dz =0. 2pi 2pi C C- Examples with solutions. Example 1. Find the residue with respect to the point at infinity for a function z+1 f (z)= . z z +1 Decision. lim =1 point z = is a removable singular point. Expression z® z 1 f (z)=1+ can be considered as its Laurent expansion in the neighborhood of an z infinitelydistant point.So c =1 and therefore resf ( )=-1 -1 . dz Example2.Calculate theintegral 1+ z4 z=2 Decision. The roots z (k =1,2,3,4) of the equation 1+ z4 =0 are the poles (finite) k ofthefunction .Note that all theseroots lieinsidethecircle z = 2. page1111 INTERNATIONAL JOURNAL OF ARTIFICIAL INTELLIGENCE ISSN:2692-5206,ImpactFactor:12,23 AmericanAcademicpublishers,volume05,issue04,2025 Journal: https://www.academicpublishers.org/journals/index.php/ijai 1 Function f (z)= in aneighborhood ofa pointat infinity has adecomposition 1+ z4 1 1 1 1 1 1 f (z)= = = - + - , 1+ z4 z4 1+ 1 z4 z8 z12 z4 that's why resf ( ) = -c =0. -1 according tothe above theorem, we have 4 resf (z )+resf ( )=0 k k=1 thisimplies dz 4 = 2pi resf (z ) = -2piresf ( ) =0. 1+ z4 k z=2 k=1 Example3.Calculate theintegral z17 dz. (z2 + 2)3(z3 +3)4 z=3 z17 Decision. The function f (z) = has five singular points z =3 (z2 + 2)3(z3 +3)4 inside the circle z , which are multiple poles. According to him, it is convenient to use the k 5 abovetheorem tocalculatethisintegral. resf (z )+resf ( )=0 k k=1 hence, z17 dz = -resf ( ) (z2 + 2)3(z3 +3)4 z=3 compute resf ( ): z17 z17 1 1 f (z) = = = . (z2 +2)3(z3 +3)4 z6(1+ 2 )2z12(1+ 3 )4 z (1+ 2 )3(1+ 3 )4 z2 z3 z2 z3 page1112 INTERNATIONAL JOURNAL OF ARTIFICIAL INTELLIGENCE ISSN:2692-5206,ImpactFactor:12,23 AmericanAcademicpublishers,volume05,issue04,2025 Journal: https://www.academicpublishers.org/journals/index.php/ijai This shows that the correct part ofthe Laurent expansion of this function inthe vicinity of the 1 pointat infinity begins with theterm . z Hence, resf ( )=-1.Thus, z17 dz = 2pi (z2 +2)3(z3 +3)4 z=3