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1、 试 题 2010 年 2011 年第 一 学期课程名称：复变函数与积分变换 专业年级： 考生学号： 考生姓名： 试卷类型： A卷 B卷 考试方式: 开卷 闭卷 一、单项选择题。（每小题3分，共15分）1若等式成立，则的值是 （ ）A B C D2设成立，则的值等于 （ ）A B C D 3下列函数中为解析函数的是 （ ）A B C D4设是函数的阶极点，那么的值为 （ ）A B C3 D5下列变换中不正确的是 （ ）AFFF BFF CFF DF二、填空题。（每小题3分，共15分）6设，则_7，则_8若C为正向圆周，则_9幂级数的收敛半径_10函数的傅里叶变换是_三、计算题。（本题6分）11
2、、求和的值四、计算下列复变函数的积分（本大题共3小题，每小题6分，共18分）12（1） （2）（3）五、解答题。（本大题共2小题，每小题6分，共12分）13试证函数是调和函数，并求函数，使得为解析函数，且满足14利用留数计算积分六、(本大题14分)15将函数在处展开为泰勒级数（6分）16将函数分别在圆环域（1）；（2）内展开为洛朗级数（8分）七、（本大题共2小题，每小题10分，共20分）17求函数的Laplace变换18利用Laplace变换求解常微分方程 2010 年 2011年第一学期 复变函数与积分变换A卷参考答案一、单项选择题。（每小题3分，共15分）1A 2 3 C 4 B二、填空题
3、。（每小题3分，共15分）6 7 2 8 0 9 10三、计算题。（本题6分）11、求和的值解： ··········································
4、;··············································（2分） ···
5、··················································
6、········································（3分） ·········&
7、#183;·················································&
8、#183;·······································（5分） ·········&#
9、183;·················································&#
10、183;········································（6分）四、计算下列复变函数的积分（本大题共3小题，每小题6分，共18分）12（1） （2）（3）解：（1）=&
11、#183;·································（4分）···············
12、83;·················································
13、83;·········································（6分）（2）·······&#
14、183;·················································&#
15、183;················（4分）································
16、3;·················································
17、3;·················（6分）（3）在曲线内，函数仅有一个奇点，且是它的一阶极点，由留数定理得：··························
18、83;·················································
19、83;·················································（2分
20、）··················································
21、;·······················（4分）··························&
22、#183;·························（6分）五、解答题。（本大题共2小题，每小题6分，共12分）13试证函数是调和函数，并求函数，使得为解析函数，且满足解：由于，显然，因此是调和函数·（1分）因为··········
23、·················································（3分）故&
24、#183;·················································&
25、#183;·····················（5分）又由得，从而·················· （6分）4利用留数计算积分解：由于函数在上半平面内只有一个极点，故 ··
26、3;·················································
27、3;······· （3分）··········································
28、;········································（5分） ·········
29、··················································
30、··································（6分）六、(本大题14分)16将函数在处展开为泰勒级数（6分）解：··········&#
31、183;·················································&#
32、183;·····························（分）····················
33、;····································（4分）·············&
34、#183;·······························（6分）17将函数分别在圆环域（1）；（2）内展开为洛朗级数（8分）解：当时············
35、;··················································
36、;············（分） ·····································&
37、#183;···············（分）当时·································&#
38、183;··（6分） ··············································
39、83;·····································（8分）七、（本大题共2小题，每小题10分，共20分）18求函数的Laplace变换解：由于 L，····&#
40、183;·················································&#
41、183;················································ （2分） 根据
42、延迟性质，得 L，··································································&#
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