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1、成都理工大学《复变函数》模拟考试试题(一)参考答案大 题一二三四总 分得 分一填空题(每题4分,共5题20分)1,则i。2三角形三个顶点复数为,则重心处的复数是。3函数的奇点是全体复数(任意复数)。4=-e。5
2、z
3、=1,化简=0。二计算题(每题6分,共4题24分)6求Ln(1+)的全体值。解:原式=···················3分=·····················3分7求的全体值。解:原式==····························3分15==[]·············3分8计算。解:
4、原式=············3分==·············3分9计算,C:的正向。解:原式=,C内只包围奇点a,····3分===··········································3分三解答题(每题6分,共4题24分)10u(x,y)=,求函数v(x,y),使u(x,y)+iv(x,y)是解析函数。解:由C-R方程,,····3分由,得15由,)=0,所以)=C。·····································3分11求级数的收敛域。解:,半径R=.······
5、·······3分圆心为。收敛域·······························3分12求映射在z=i处的伸缩率和转动角。解:,·························3分。伸缩率为2,转动角。·······3分13在映射w=(1+i)下,z平面上的图形
6、z-i
7、<1被映射成w平面上的什么图形。解:共轭表示以实轴为对称轴上下翻,映射为
8、z+i
9、<1·····3分,乘以(1+i)表示旋转,再以原点为基准向外膨胀为倍的圆。15圆心是,膨胀到1-i。半径从1膨胀到。最后图形是
10、w-(1-i)
11、<。······
12、······················3分四计算题(每题8分,共4题32分)14计算,L是连接(0,0)与(,0)的曲线,(0,0)为起点。解:原式=··············定义2分=·····展开2分=·············代路径2分=·······································分部积分2分15把函数在区域1<
13、z
14、<2展开成洛朗级数。解:原式=····拆项2分其中·····················2分···························2分15···
15、·····················2分16判断函数的所有有限奇点与无穷远奇点的类型,并计算每个奇点的留数。解:有限奇点有z=0,z=2,分别是一级极点。··············2分因为,所以z=是可去奇点。······2分`········两个有限奇点的留数2分留数定理2:·····2分17求将下半平面Im(z)<0映射成单位圆
16、w-i
17、<2的分式线性映射。解:下半平面取任一复数。映射到0,映射到。,·········································2分当z=1,
18、w
19、=1,所以
20、k
21、
22、=1.下半平面映射到单位圆内,··········································2分扩大为半径为2的圆内15··········································2分往上平移至圆心为i.+i·········································2分15成都理工大学《复变函数》模拟考试试题(二)附参考答案大 题一二三四总 分得 分一.判断题(每题2分,共6题,总分12分)1.若是和的一个奇点,则也是和的奇点(╳)2.若函数点解析,则在该
23、点连续。(√)3. (√)4.,,则是解析函数(╳)5.若存在,则与有相同的收敛半径(√)6.是的孤立奇点(╳)二.填空题(每题3分,共8题,总分24分)1.设,则2.设,则的指数形式表示为3.函数将平面上的曲线变为平面上的曲线的方程是(用u,v表示)。4.151.-12.,则。3.映射在z=i处转动角是0,伸缩率是ch14.函数的奇点是所有纯虚数,(y轴上除0以外的所有点)一.解答题(每题6分,共8题,总分48分)1.证明:解:左边=····················
24、2分······················2分·····················2分2.解方程解:·····················2分·······································2分··················