hosila moduli va argumentining geomertik ma’nosi conform akslantirishlar

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1526023764_71481.doc z c d ì ) ( z f w = d z î 0 ) 0 ) ( ' ( ) ( ' 0 0 ¹ z f z f 0 0 0 0 0 0 0 lim ) ( ) ( lim ) ( ' z z w w z z z f z f z f z z z z - - = - - = ® ® ( ) ) ( 0 0 z f w = ( ) 0 0 0 0 0 ) ( ' z z z z z f w w - + - = - 0 z z - 0 z z - 0 w w - ) ( ' 0 z f ) ( z f w = 0 zzr -= ( ) 0 0 z z - 00 '() wwfzr -=× 1 ) ( ' 0 z f ) …

sh nuqtaning biror atrofida hosilaga ega bo’lib bo’lsin. nuqtadan o’tuvchi silliq egri chiziqni olib, uning yo’nalishi bo’yicha shu egri chiziqqa nuqtada urinma o’tkazamiz bu urinmaning haqiqiy o’kning musbat qismi bilan tashkil etgan burchagi bo’lsin. akslantirish esa egri chiziqni tekislikda g egri chiziqqa o’tkazsin. murakkab funksiyaning hosilasini hisoblash qoidasiga binoan bo’lib, t=t0 da embed equation.3 bo’ladi. shartga ko’ra va ( ning silliqligidan) bo’lgani uchun bo’ladi. binobarin, nuqtada egri chiziqning urinmasi mavjud. bu urinmaning burchak koeffitsentini bilan belgilaymiz: tenglikdan ya’ni kelib chiqadi. agar mikdorning akslantirish natijasida egri chiziqning nuqtadagi burilishi burchagi ekanligini e’tiborga olsak, u holda (2) tenglikdan nuqtadan o’tuvchi barcha silliq egri chiziqlar bir xil burchakka burilishini ko’ramiz (burchakning saqlanishi). ta’rif: agar akslantirish nuqtada cho’zilish va burchak saqlanish xossalariga ega bo’lsa, bunday akslantirishga nuqtada konform akslantirish deyiladi. yuqoridagilardan ko’rinadiki, agar funksiya nuqtaning biror atrofida golomorf bo’lib, bo’lsa, akslantirish nuqtada konform bo’ladi. agar akslantirish d sohada bir yaproqli bo’lib, sohaning har bir …

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1526023764_71481.doc z c d ì ) ( z f w = d z î 0 ) 0 ) ( ' ( ) ( ' 0 0 ¹ z f z f 0 0 0 0 0 0 0 lim ) ( ) ( lim ) ( ' z z w w z z z f z f z f z z z z - - = - - = ® ® ( ) ) ( 0 0 z f w = ( ) 0 0 0 0 0 ) ( ' z z z z z f w w - + - = - 0 z z - 0 z z - 0 w w - ) ( ' 0 …

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