复变函数与积分变换A卷参考答案(2014年12月)

注：1、教师命题时题目之间不留空白；2、考生不得在试题纸上答题，教师只批阅答题册正面部分，若考生须在试题图上作解答，请另附该试题图。3、请在试卷类型、考试方式后打勾注明。（第1页）一、单项选择题。（每小题3分，共18分）1．复数13zi的辐角主值为(C)A.3B.3C.3D.+3k2．13zei，则Re()z为(A)A．2lnB．2ln2C．3D．233．设22fzxyix，那么(B)A.fz在复平面上处处可微B.fz仅在x轴上可导C.fz仅在原点可导D.fz在复平面上处处不可导4．z平面上的直线2x经(1)2uivizi映射后在平面上的图形是(A)A．50uvB．50uvC．50uvD．50uv5．0z是函数31)(zezfz的(C)A．可去奇点B．本性奇点C．二阶极点D．三阶极点6．()sintt(C)A．0B．1C．sintD．tcos二、填空题。（每小题3分，共18分）7．2lim(1)zizizz12．8．设C为正向圆周1z，则zCzedz0．9．幂级数0(1)nnnzi的收敛半径为2.10．若幂级数0()nnnczi在点2zi处发散，则该级数在点2z处的敛散性为发散．注：1、教师命题时题目之间不留空白；2、考生不得在试题纸上答题，教师只批阅答题册正面部分，若考生须在试题图上作解答，请另附该试题图。3、请在试卷类型、考试方式后打勾注明。（第2页）11．单位阶跃函数()ut的Laplace变换为1s．12．224()(+4)Fsss的Laplace逆变换为1(sin2)()2ttut．三、计算题。（每小题8分，共16分）13．设复数34(1)izi，（1）求z的实部和虚部；（2）求z的模.解：341+(1)izii·················································································（4分）(1)Re1z,Im1z·············································································（6分）(2)2z·······························································································（8分）14．利用留数计算积分222(1)(4)xdxxx.解：函数222()(1)(4)zfzzz在上半平面内有两个一阶级点i，2i，························（2分）又222Re[(),]lim()(1)(4)6zizisRzizizz··················································（４分）2222Re[(),2]lim(2)(1)(4)3zizisRzizizz···········································（6分）因此，222(1)(4)xdxxx=2{Re[(),]Re[(),2]}isRzisRzi=2()633iii（8分）四、计算题。（每小题8分，共16分）15．设C为正向圆周2z，计算积分1(1)zCeIdzzz．注：1、教师命题时题目之间不留空白；2、考生不得在试题纸上答题，教师只批阅答题册正面部分，若考生须在试题图上作解答，请另附该试题图。3、请在试卷类型、考试方式后打勾注明。（第3页）解：1111()11zzzzCCCeeeeIdzdzdzzzzz···········································（4分）012(1)2(1)zzzzieie-·······························································（6分）12(1)ei·························································································（8分）16．设C表示正向圆周3，2371()Cfzdz，求(1)fi.解根据柯西积分公式知，当z在c内时，22()2(371)2(371)zfziizz·························（6分）故()2(67)fziz，而1i在c内，所以(1)2(613)fii.························（8分）五、解答题。（每小题8分，共16分）17．设),(),()(yxivyxuzf是解析函数，其中(,)4uxyxyy，且(0)fi，求),(yxv.解：由C-R条件，有41vuxxy，4vuyyx······································（２分）则()4(41)uvfziyixxx4()4ixyiiizi··································（４分）∴2()()2fzfzdzizizC·······································································（６分）又∵(0)0f∴Ci故2()(21)fzizz················································（８分）18．求21(1)zz在圆环域11z内的洛朗级数展开式．解：101111(1)(1)11(1)111nnnzzzzz--==，···········································(４分)1220111[(1)(1)](1)(1)(1)nnnnnnznzzz-·······························（6分）注：1、教师命题时题目之间不留空白；2、考生不得在试题纸上答题，教师只批阅答题册正面部分，若考生须在试题图上作解答，请另附该试题图。3、请在试卷类型、考试方式后打勾注明。（第4页）3211(1)(1)(1)(1)nnnnzzz···································································（８分）六、解答题。（每小题8分，共16分）19．利用Laplace变换求微分方程tyye满足初始条件(0)1y的特解．解：原方程两边取Laplace变换，1()(0)()1sYsyYss--，···································（4分）又(0)1y，因此2()(1)sYss，····································································（6分）从而方程的解为2221()Re,1lim[(1)](1),0(1)(1)ststtsseseytsstetss.·······（8分）20．已知224[]tee，利用Fourier变换的性质求函数22()tftte的Fourier变换．解：由导数的像函数公式，得224[2]tteei，···········································（4分）又由像函数的导数公式,得2222244[2]()(1)2tteieie，·············（6分）因此22224[](1)22ttee········································································（8分）