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OLIY TA'LIM MUASSASALARI TALABALARIGA KOSHINING INTEGRAL FORMULASI MAVZUSINI TUSHUNTIRISH METODIKASI
Sabohat Dusimbatovna Eshmetova
Chirchiq davlat pedagogika universiteti "Matematika va informatika" fakulteti
o'qituvchisi
ANNOTATSIYA
Kompleks o'zgaruvchili funksiya integralining yechimi ushbu maqolada Koshining integral formulasidan foydalanib bayon etilgan.
Kalit so'zlar: Golomorf funksiyalar, Koshining integral formulasi, bog'lamli
soha.
ABSTRACT
The solution to the Integral of a complex variable function is stated in this article using Koshy's integral formula
Keywords: Holomorphic functions, Cauchy's integral formula is, connected
field.
Kompleks sonlar tekisligi Cz da chegaralangan D sohani qaraylik. Uning chegarasi dD silliq (bo'lakli silliq) chiziqdan iborat. Bu yopiq egri chiziq musbat yo'nalishda olingan bo'lsin:
Aytaylik, D = D ^>sd to'plamda f (z) funksiya aniqlangan bo'lsin.
1-teorema ([2],[4],). Agar f (z) funksiya D sohada golomorf bo'lib, D da uzluksiz bo'lsa, u holda Vz e D nuqta uchun
1 rf (t)
f (z) = ^ i 7
2m Dt
dt
- z
tenglik o'rinli bo'ladi.
2-teorema ([3],[4],[5]). Koshi tipidagi integral C \ r sohada F(z) funksiyasini aniqlab, bu funksiya ushbu xossalarga egadir:
a) F(z) funksiya C \ T da golomorf,
b) lim F (z) = 0
z—
c) F(z) funksiyaning istalgan tartibli hosilasi Fn(z) mavjud va
n! r f (f)
n' (•
Fn (z) = — f
2m i (f- z)
n+1
df
(1)
tenglik o'rinli. https://cspi.uz/
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1- misol. Ushbu
i
2 z
(z + 3)4
-dz
integralni hisoblang, bunda y chiziq C tekislikdagi z = — 3 nuqtani o„z ichiga oladigan ixtiyoriy yopiq kontur.
y kontur bilan chegaralangan sohani D deb belgilaymiz. Ravshanki, f(z) = e 2 z funksiya va D soha uchun 2-teoremaning shartlari bajariladi. Unda (1)-formuladan foydalanib topamiz:
dz = dz = ™• r (-3) = — • 23 • e^ = ^. J (z + 3)4 ^ ■ f ( ) " °~6
(z + 3)4
3!
3e
2- misol. Koshining integral formulasidan foydalanib quyidagi
i
ez dz
-2.5( z + 4)( z - 6)
integralni hisoblang.
|z — 2| =5 aylana bilan chegaralangan sohani D deb belgilaymiz.
F (z) =
(z + 4)( z - 6)
deb belgilasak, z x = 6 e D va z 2 = — 4 g D
f (z)
^ F (z) = —
(z + 4)( z - 6) z - 6
f(z) funksiya D da golomorf bo'lganligi uchun Koshining integral formulasiga muvofiq
r ez dz
i F(z)dz = i f(z)dz = 2xif(6) = 2rn —
. e36 _ e26m
z - 6
10 5
|z-21=5 (z + 4)(z - 6) |z-2|=5 |z-2=5-
3- misol. Koshining integral formulasidan foydalanib quyidagi
f z -1 dz
M=U z + 2)( z - 3)
integralni hisoblang.
|z| = 2 , 5 aylana bilan chegaralangan sohani D deb belgilaymiz.
z -1
F ( z) =
(z + 2)( z - 3) z-1
deb belgilasak, z x = — 2 e D , z 2 = 3 g D .
f ( z)
^ F (z) = -
(z + 2)( z - 3) z + 2 f(z) funksiya D da golomorf bo'lganligi uchun Koshining integral formulasiga muvofiq
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October 20, 2023 Republican Scientific and Practical Conference
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z -1
-dz = f I J
f (z)dz „ ... „. „ . 3 6mi J w = 2mf (-2) = 2m ■- = ■
z + 2
5 5
И=2,5(z + 2)(z - 3) W=2,5
4- misol. Koshining integral formulasidan foydalanib quyidagi
г z -1
И=з(z - 3)2(z +i)2 integralni hisoblang.
|z + 1| =3 aylana bilan chegaralangan sohani D deb belgilaymiz.
z -1
dz
F ( z ) =
(z - 3)2( z + i)2 z-1
deb belgilasak, zt = —i £ D, z2 = 3 ^ D.
_ f (z)
^ F (z) = - , ,
V ' (z - 3)2(z + i)2 (z + i)2 f(z) funksiya D da golomorf bo'lganligi uchun Koshining integral formulasiga muvofiq
z -1
|z+:
i
1=3(z - 3)2( z +i)2 |z+-^1=3( z + i)
dz = i dz = 2mi ■ f '(-0 = 2m(i +1)
(i + 3)3
REFERENCES
1. Sadullayev A., Xudoyberganov G., Mansurov X., Vorisov A., Tuychiyev T.
2. Matematik analiz kursidan misol va masalalar to'plami. 3-qism (kompleks analiz) "O'zbekiston", 2000.
3. Волковыский Л.И., Лунц Г.Л., Араманович И.Г. Сборник задач по теории функций комплексного пременного. 3-nashri. - М. "Наука", 1975.
4. Xudoyberganov G., Vorisov A., Mansurov X. Kompleks analiz. (ma'ruzalar). T, "Universitet", 1998.
5. Шабат Б.В. Введение в комплексный анализ. 2-nashri, 1-ч.-М, "Наука", 1976.
6. E. M. Mahkamov, S. D. Eshmetova. Chegirmalar yordamida xosmas integrallarni hisoblash usullari// academic research in educational sciences volume 2 | ISSUE 9 | 2021 p. 91-100
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October 20, 2023 Republican Scientific and Practical Conference
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