第 1页 /共 1页工程数学(考试形式： 闭卷 考试时间: 2小时)考试作弊不授予学士学位方向： 姓名： ______ 学号： ______1. Find values of:(a) );3(Ln − (b) )i +(12.(10 points)2. Function is harmonic, find an analytic functionsuch that satisfying (0)0f = .(10 points)3. Evaluate each of the following integrals: (20 points) 22;(9)()z zz z z i −+∫(b) d23131(2)z z z z −=−∫ (d)d .4. Find the series representation for the function at .(10 points)5. Evaluate integral of , where . (10 points)6. Find a representation for the function in powers of .(10 points)7. Find the residue of function 6sin ()z z f z z−=at 0z =.(10 points)8. Find the inverse Laplace transform of function 225()(2)9s F s s +=++. (10 points)9. Evaluate integral along positively oriented circle . (10 points) 2(1)z z e z z z =−∫2(a)d ; 10||2()(1)(3)z z z i z z =+−−∫d (c); (,)(cos sin ),()x v x y e y y x y x y f z u iv =+++=+ arctan 0z z = 2sin 14112Cz z C z z π+=−∫d : 11ze z − 1:|-2|2z iCdz C z eiππ=−∫第 1页 /共 3页《工程数学》期末试题答案(B)1.(a) (5 points)1.(b) (5 points)2.(10 points) 3.(a) z=0为一级极点, z=1二级极点(5 points)(b) (5 points))2sin(ln )2[cos(ln 2 0 .,2,1,0 )],2sin(ln )2[cos(ln 2)]22sin(ln )22[cos(ln 2222ln )22(ln )22(ln ) 2ln2)(1(2Ln )1(1i k k i e k i k e e e e k k k i k i k i i i +=±±=+=+++====−−++−++++时，得其主值为其中L πππππππ),2,1,0(,)12(3ln )3(Arg 3ln )3(Ln L ±±=++=−+−=−k i k i 其中π,1)sin sin cos (+++=∂∂y y x y y e xv x ,1)cos sin (cos ++−=∂∂y x y y y e y v x,1)cos sin (cos ++−=∂∂=∂∂y x y y y e y v x u x 由),()sin cos (d ]1)cos sin (cos [ y g x y y y x e x y x y y y e u x x ++−=++−=∫得 , 得由y u xv ∂∂−=∂∂),()sin cos sin (1)sin sin cos (y g y y y y x e y y x y y e x x ′−++=+++,)( C y y g +−=故,)sin cos ( C y x y y y x e u x+−+−=于是,)1()1()1()(C z i ze C i iy i x e iye e xe iv u z f z iy x iy x +++=++++++=+= ,0)0( =f 由,0 =C 得.)1()( z i ze z f z ++=所求解析函数为z z z e z z f z z d )1(lim ]0),([Res 20−⋅=→,1)1(lim 20=−=→z e zz ⎥⎦⎤⎢⎣⎡−−−=→221)1()1(d d lim )!12(1]1),(Res[z z e z z z f z z ⎟⎟⎠⎞⎜⎜⎝⎛=→z e z z z d d lim 10)1(lim 21=−=→z z e z z z z z e C z d )1(2∫−{}]1),(Res[]0),(Res[2z f z f i +=π.2i π=∫=+−22d ))(9(z z i z z z .592d )(9222ππ=−⋅=−−−=−==∫i z z z z i z i z z z第 2页 /共 3页(c)由于－i 与1在C 内部，(5 points) (d)2233131132|(2)8z z d idz i z z dz z ππ=−=−==−∫(5 points) 4.(10 points)5.(10 points)6.(10 points)2, 23 ,0 2 )2(132==−===−z z C z z z z 仅包含奇点和有两个奇点函数;2214sin 2d 114sin d 14sin 12112112i z zi z z z zz z z z z z πππππ=−⋅=+−=−−==+=+∫∫,1d arctan 02∫+=z z z z 因为1,)()1(11 022<⋅−=+∑∞=z z z n nn 且∫+=z z z z 021d arctan 所以∫∑∞=⋅−=z n n n z z 002d )()1(.1,12)1(012<+−=∑∞=+z n z n n ni,1,3)3)(1()(1)(10−∞−−+=点外，其他奇点为除被积函数z z i z z f 0]),(Res[]3),(Res[]1),(Res[]),(Res[ =∞+++−z f z f z f i z f 则∫−−+Cz z i z z )3)(1()(d 10]}1),(Res[]),(Res[{2z f i z f i +−=π]}),(Res[]3),(Res[{2∞+−=z f z f i π.)3(0)3(2121010i i i i +−=⎭⎬⎫⎩⎨⎧++−=ππ211)1(1)(z e z f z −=′−,)1(1)(2z z f −=,0)()()1( 2=−′−z f z f z 所以0)()32()()1(2=′−+′′−z f z z f z 0)(2)()54()()1(2=′+′′−+′′′−z f z f z z f z L L L ,13)0(,3)0(,)0()0(e f e f e f f =′′′=′′=′=).1(,!313!2313211<⎟⎠⎞⎜⎝⎛++++=−z z z z e e z L第 3页 /共 3页7.利用洛朗展开式(10 points) 8.(10 points)9．由)22(ππk iLnii e e i +−==可知被积函数11)(−=z e z f 以,...)2,1,0(),22(±±=+−=k k z k ππ为一阶极点，其中)42(),22(21ππππ+−=+−=−−z z 包含在ππ2||=−z 内部，由公式,...)2,1,0(|)'(1]),([Re 22++==−=+−k e i e z z f s k z z i z k k ππ，由留数定理，)(2]}),([Re ]),([Re {2)(12723212|2|ππππππ−−−−=−+=+=−∫ee i z zf s z z f s i i e z i z(10 points)223)2(1)2(2)(++++=s s s F )3sin 313cos 2(]}31[]3[2{]312[]3)2(1)2(2[)]([2221221222122211t t e s L s s L e s s L e s s L s F L tt t +=+++=++=++++=−−−−−−−−(0)(0)(0)0,P P P ′′′===(0)0.P ′′′≠3566sin 13!5!z z z z z z z z ⎡⎤⎛⎞−=−−+−⎢⎥⎜⎟⎝⎠⎣⎦L 16sin 1,0.5!z z c z −−⎡⎤∴==−⎢⎥⎣⎦Res。
