Đề thi môn Hàm phức và phép biến đổi Laplace - Đề số 4

= D. ( )( ) cosRe xxe y f z y = − Câu 2: Khai triển Laurent của hàm ( ) ( ) 12 1 cosf z z z   = +     trong lân cận của điểm z = 0 là: A. ( ) ( ) ( ) 20 2 1 1 1 2 2 ! 2 ! n n n zn n ∞ =   − +   +   ∑ B. ( ) ( ) ( ) 20 2 1 1 1 2 2 ! 2 ! n n n zn n ∞ =   − +   +   ∑ C. ( ) ( ) 22 10 2 1 1 !1 ! n nn n n zn z ∞ − =   − +   +   ∑ D. ( ) ( ) ( )2 1 20 2 1 1 2 ! 2 ! n n n n n z n z ∞ − =   − +     ∑ Câu 3: Cho hàm phức ( ) ( ) 3 2 6 18 ze f z z z z = + + . Hãy chọn phát biểu SAI: A. 3 3z i=− + là cực điểm cấp 1 B. 0z = là cực điểm cấp 2 C. 3 3z i= − − là cực điểm cấp 1 D. 3 3z i=− + và 3 3z i= − − là các điểm bất thường cô lập Câu 4: Giả sử hàm gốc ( )f t có ảnh là ( )F p , ( ) ( )f t F p  =  L . Hãy chọn phát biểu ĐÚNG: A. ( )3 3 t pe f t F    =        L B. ( ) ( )3 3 *t F p e f t p −  =  L C. ( ) ( )3 0 3 3 t u F p e f u dt p   −  =  −   ∫L D. ( ) ( )3 3te f t F p  = −  L Câu 5: Tìm ảnh của hàm gốc ( )2 0 * sin 3 t te u du∫ : A. ( )2 1 3 2 9p p p + − + B. 2 1 1 3 2 9p p p + + − + C. ( )( )2 3 2 9p p p− + D. ( )( )2 3 2 9p p− + Câu 6: Tìm biến đổi Laplace ( )2 sin 5tte t−   L : Trang 2/6 - Mã đề thi 1001-060-485 A. ( ) ( ) 2 2 2 10 20 sin 5 4 29 t pte t p p − +  =   + + L B. ( ) ( ) 2 2 2 10 20 sin 5 2 25 t pte t p − −  =     + +    L C. ( ) ( ) ( ) 2 2 2 10 2 sin 5 2 25 t p te t p − +  =     − +    L D. ( ) ( ) 2 2 2 10 20 sin 5 4 29 t pte t p p − −  =   − + L Câu 7: Cho hàm phức ( ) ( )2 sin 2 1 z f z z z π = − . Hãy chọn phát biểu ĐÚNG: A. ( )( )Res ,0f z iπ= − và ( ) 1Res , 2 2 f z    =    B. ( )( )Res ,0f z π= − và ( ) 1Res , 2 2 f z    =    C. ( )( )Res ,0 2f z = và ( ) 1Res , 2 f z π    = −    D. ( )( )Res ,0f z π= − và ( ) 1Res , 4 2 f z    =    Câu 8: Cho hàm ( )f z có khai triển Laurent tại trong lân cận của điểm z = 0 là: ( ) ( ) ( ) ( ) 2 2 1 2 0 2 1 1 . 2 ! 2 ! n n n n n f z n z n z ∞ + =   = − +     ∑ Tính tích phân ( )5 | | 2 . z I z f z dz = = ∫
A. 2 6! iπ − B. 4 12 5! 6! iπ   −    C. 2 6! iπ D. 8 5! iπ Câu 9: Cho hàm số ( ) ( ), cos .xu x y ax e ay= + Xác định hằng số phức a sao cho ( , )u x y là phần thực của một hàm giải tích trên ℂ . A. 0a = B. 1a = hoặc 1a = − C. 1a = hoặc 2a = D. Không tồn tại a Câu 10: Biến đổi Laplace ngược nào sau đây là SAI: A. 1 2 2 1 3 2 t te e p p −     = − − +  L B. 3 1 2 2 3 2 3 1 2 3 t te e p p −    − = − − +  L C. ( ) ( ) ( )1 2 2 1 1 2cos 2 sin 2 21 4 tp e t t p −     −   = +    − +   L D. ( ) ( )1 2 3 2 2 3cos 3 sin 3 39 p t t p −  −  = − +  L -- ----------------------------------- PHẦN TỰ LUẬN (5 ĐIỂM) Câu 11 (1.5 điểm). Áp dụng phép biến đổi Laplace giải phương trình vi phân sau: 1ty y te′′ + = + với điều kiện ( ) ( )0 0 0.y y′= = Câu 12 (2.0 điểm). Áp dụng phép biến đổi Laplace giải phương trình tích phân: ( )2 2 0 * . t t ty e y u du t e+ = +∫ Câu 13 (1.5 điểm). Cho hàm phức ( ) 3 1 . zf z ze −= a) Khai triển Laurent hàm f trong lân cận của điểm 1.z = b) Sử dụng kết quả này tính tích phân ( ) | | 3 . z i I f z dz − = = ∫
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Trang 6/6 - Mã đề thi 1001-060-485 TRƯỜNG ĐH SƯ PHẠM KỸ THUẬT TPHCM KHOA KHOA HỌC CƠ BẢN BỘ MÔN TOÁN ĐỀ THI CUỐI KỲ HỌC KỲ II NĂM HỌC 2014-2015 Môn: HÀM PHỨC VÀ PHÉP BIẾN ĐỔI LAPLACE Mã môn học: 1001060 Đề số/Mã đề: 1001-060-485 Thời gian: 75 phút Được phép sử dụng tài liệu Chữ ký giám thị 1 Chữ ký giám thị 2 CB chấm thi thứ nhất CB chấm thi thứ hai Điểm và chữ ký Điểm và chữ ký Họ và tên: .................................................................... Mã số SV: .................................................................... Số TT:........................ Phòng thi:................................ PHẦN TRẢ LỜI CÂU HỎI TRẮC NGHIỆM CÂU HỎI 1 2 3 4 5 6 7 8 9 10 TRẢ LỜI PHẦN TRẢ LỜI CÂU HỎI TỰ LUẬN ......................................................................................................................................................................... ......................................................................................................................................................................... ......................................................................................................................................................................... ......................................................................................................................................................................... ......................................................................................................................................................................... ......................................................................................................................................................................... ......................................................................................................................................................................... ......................................................................................................................................................................... ......................................................................................................................................................................... ......................................................................................................................................................................... ......................................................................................................................................................................... ......................................................................................................................................................................... ......................................................................................................................................................................... ......................................................................................................................................................................... ......................................................................................................................................................................... ......................................................................................................................................................................... ......................................................................................................................................................................... ......................................................................................................................................................................... ......................................................................................................................................................................... ......................................................................................................................................................................... ......................................................................................................................................................................... .........................................................................................................................................................................