复变函数与积分变换(修订版-复旦大学)第六章课后的习题答案-(1).pdf

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复变函数与积分变换(修订版-复旦大学)第六章课后的习题答案-(1).pdf

精品资料欢迎下载习题六1.求映射1wz下，下列曲线的像.(1)22xyax (0a，为实数)解：222211i=+iixywuvzxyxyxy221xxuxyaxa,所以1wz将22xyax 映成直线1ua.(2).ykx(k为实数)解：22221ixywzxyxy222222xykxuvxyxyxyvku故1wz将 ykx 映成直线 vku.2.下列区域在指定的映射下映成什么？（1）Im()0,(1i)zwz；解：(1i)(i)()i(+)wxyxyx y,.20.uxy vxyuvy所以 Im()Re()ww.故(1i)wz 将 Im()0,z映成 Im()Re()ww.(2)Re(z)0.0Im(z)0,0y0.Im(w)0.若w=u+iv,则2222,uvyxuvuv因为 0y0,0Im(z)0,Im(w)0,1212w (以（12,0）为圆心、12为半径的圆)-第 1 页，共 7 页精品p d f 资料可编辑资料-精品资料欢迎下载3.求w=z2在z=i 处的伸缩率和旋转角，问w=z2将经过点z=i 且平行于实轴正向的曲线的切线方向映成w平面上哪一个方向？并作图.解：因为 w=2z,所以 w(i)=2i,|w|=2,旋转角 arg w=2.于是,经过点 i 且平行实轴正向的向量映成w平面上过点-1，且方向垂直向上的向量.如图所示.4.一个解析函数，所构成的映射在什么条件下具有伸缩率和旋转角的不变性？映射w=z2在z平面上每一点都具有这个性质吗？答：一个解析函数所构成的映射在导数不为零的条件下具有伸缩率和旋转不变性映射w=z2在z=0 处导数为零，所以在z=0处不具备这个性质.5.求将区域0 x0.解：(1)Re(z)=0 是虚轴，即z=iy代入得.22222i1(1i)12ii1111yyyywyyyy写成参数方程为2211yuy，221yvy,y.消去y得，像曲线方程为单位圆,即u2+v2=1.(2)|z|=2.是一圆围，令i2e,02z.代入得ii2e12e1w化为参数方程.354cosu4sin54cosu02消去得，像曲线方程为一阿波罗斯圆.即22254()()33uv(3)当 Im(z)0 时，即11Im()011wwzww,令w=u+iv得221(1)i2Im()Im()01(1)i(1)wuvvwuvuv.即v0,故 Im(z)0 的像为 Im(w)0.9.求出一个将右半平面Re(z)0 映射成单位圆|w|0,映射成|w|0,映为单位圆|w|0).(1)由f(i)=0得=i，又由 arg(i)0f,即i22i()e(i)fzz,i()21(i)e02f，得2，所以iiizwz.(2)由f(1)=1,得k=11；由f(i)=15,得k=i5(i)联立解得3+(52i)(52i)3zwz.12.求将|z|1 映射成|w|1 的分式线性变换w=f(z)，并满足条件：(1)f(12)=0,f(-1)=1.(2)f(12)=0,12arg()2f,(3)f(a)=a,arg()fa.解：将单位圆|z|1 映成单位圆|w|1 的分式线性映射，为ie1zwz ,|1.(1)由f(12)=0，知12.又由f(-1)=1，知1iii2121ee(1)1e11.故12221112zzzwz.-第 4 页，共 7 页精品p d f 资料可编辑资料-文档编码:CG5K2J3P2E3 HK3X4G5S3X4 ZO8C7C4A10P5文档编码:CG5K2J3P2E3 HK3X4G5S3X4 ZO8C7C4A10P5文档编码:CG5K2J3P2E3 HK3X4G5S3X4 ZO8C7C4A10P5文档编码:CG5K2J3P2E3 HK3X4G5S3X4 ZO8C7C4A10P5文档编码:CG5K2J3P2E3 HK3X4G5S3X4 ZO8C7C4A10P5文档编码:CG5K2J3P2E3 HK3X4G5S3X4 ZO8C7C4A10P5文档编码:CG5K2J3P2E3 HK3X4G5S3X4 ZO8C7C4A10P5文档编码:CG5K2J3P2E3 HK3X4G5S3X4 ZO8C7C4A10P5文档编码:CG5K2J3P2E3 HK3X4G5S3X4 ZO8C7C4A10P5文档编码:CG5K2J3P2E3 HK3X4G5S3X4 ZO8C7C4A10P5文档编码:CG5K2J3P2E3 HK3X4G5S3X4 ZO8C7C4A10P5文档编码:CG5K2J3P2E3 HK3X4G5S3X4 ZO8C7C4A10P5文档编码:CG5K2J3P2E3 HK3X4G5S3X4 ZO8C7C4A10P5文档编码:CG5K2J3P2E3 HK3X4G5S3X4 ZO8C7C4A10P5文档编码:CG5K2J3P2E3 HK3X4G5S3X4 ZO8C7C4A10P5文档编码:CG5K2J3P2E3 HK3X4G5S3X4 ZO8C7C4A10P5文档编码:CG5K2J3P2E3 HK3X4G5S3X4 ZO8C7C4A10P5文档编码:CG5K2J3P2E3 HK3X4G5S3X4 ZO8C7C4A10P5文档编码:CG5K2J3P2E3 HK3X4G5S3X4 ZO8C7C4A10P5文档编码:CG5K2J3P2E3 HK3X4G5S3X4 ZO8C7C4A10P5文档编码:CG5K2J3P2E3 HK3X4G5S3X4 ZO8C7C4A10P5文档编码:CG5K2J3P2E3 HK3X4G5S3X4 ZO8C7C4A10P5文档编码:CG5K2J3P2E3 HK3X4G5S3X4 ZO8C7C4A10P5文档编码:CG5K2J3P2E3 HK3X4G5S3X4 ZO8C7C4A10P5文档编码:CG5K2J3P2E3 HK3X4G5S3X4 ZO8C7C4A10P5文档编码:CG5K2J3P2E3 HK3X4G5S3X4 ZO8C7C4A10P5文档编码:CG5K2J3P2E3 HK3X4G5S3X4 ZO8C7C4A10P5文档编码:CG5K2J3P2E3 HK3X4G5S3X4 ZO8C7C4A10P5文档编码:CG5K2J3P2E3 HK3X4G5S3X4 ZO8C7C4A10P5文档编码:CG5K2J3P2E3 HK3X4G5S3X4 ZO8C7C4A10P5文档编码:CG5K2J3P2E3 HK3X4G5S3X4 ZO8C7C4A10P5文档编码:CG5K2J3P2E3 HK3X4G5S3X4 ZO8C7C4A10P5文档编码:CG5K2J3P2E3 HK3X4G5S3X4 ZO8C7C4A10P5文档编码:CG5K2J3P2E3 HK3X4G5S3X4 ZO8C7C4A10P5文档编码:CG5K2J3P2E3 HK3X4G5S3X4 ZO8C7C4A10P5文档编码:CG5K2J3P2E3 HK3X4G5S3X4 ZO8C7C4A10P5文档编码:CG5K2J3P2E3 HK3X4G5S3X4 ZO8C7C4A10P5文档编码:CG5K2J3P2E3 HK3X4G5S3X4 ZO8C7C4A10P5文档编码:CG5K2J3P2E3 HK3X4G5S3X4 ZO8C7C4A10P5文档编码:CG5K2J3P2E3 HK3X4G5S3X4 ZO8C7C4A10P5文档编码:CG5K2J3P2E3 HK3X4G5S3X4 ZO8C7C4A10P5文档编码:CG5K2J3P2E3 HK3X4G5S3X4 ZO8C7C4A10P5文档编码:CG5K2J3P2E3 HK3X4G5S3X4 ZO8C7C4A10P5文档编码:CG5K2J3P2E3 HK3X4G5S3X4 ZO8C7C4A10P5文档编码:CG5K2J3P2E3 HK3X4G5S3X4 ZO8C7C4A10P5文档编码:CG5K2J3P2E3 HK3X4G5S3X4 ZO8C7C4A10P5文档编码:CG5K2J3P2E3 HK3X4G5S3X4 ZO8C7C4A10P5文档编码:CG5K2J3P2E3 HK3X4G5S3X4 ZO8C7C4A10P5精品资料欢迎下载(2)由f(12)=0，知12，又i254e(2)zwzi11224()earg()32ff，于是21i2221e()i12zzzwz.(3)先求=()z，使z=a0，arg()a,且|z|1 映成|1.则可知i=()=e1zaza z再求w=g()，使=0w=a,arg(0)0g,且|1 映成|w|1.先求其反函数=()w,它使|w|1 映为|1,w=a映为=0，且arg()arg(1/(0)0wg,则=()=1wawa w.因此，所求w由等式给出.i=e11wazaa wa z.13.求将顶点在0，1，i 的三角形式的内部映射为顶点依次为0，2，1+i 的三角形的内部的分式线性映射.解：直接用交比不变性公式即可求得02ww1i01i2=02zzi0i12ww.1i21i=1zz.i1i4z(i1)(1i)wz.14.求出将圆环域2|z|5 映射为圆环域4|w|2 映为|w|10.又w=f(z)将|z|=5 映为|w|=4，将z=2映为w=-10，所以将|z|4，由此确认，此函数合乎要求.15.映射2wz将z平面上的曲线221124xy映射到w平面上的什么曲线？解：略.16.映射w=ez将下列区域映为什么图形.(1)直线网 Re(z)=C1,Im(z)=C2;(2)带形区域Im(),02z;(3)半带形区域Re()0,0Im(),02zz.解：（1）令z=x+iy,Re(z)=C1,z=C1+iy1i=eeCyw,Im(z)=C2,则z=x+iC22i=eeCxw故=ezw将直线 Re(z)映成圆周1eC；直线 Im(z)=C2映为射线2C.（2）令z=x+iy，y,则ii=eeee,zxyxywy故=ezw将带形区域Im()z映为arg()w的张角为的角形区域.（3）令z=x+iy，x0，0y0,0Im(z)1,0arg w(02).17.求将单位圆的外部|z|1 保形映射为全平面除去线段-1Re(w)1 映为|w1|1,再用分式线性映射.1211i1www将|w1|0,然后用幂函数232ww映为有割痕为正实轴的全平面，最后用分式线性映射3311www将区域映为有割痕-1,1的全平面.-第 6 页，共 7 页精品p d f 资料可编辑资料-文档编码:CG5K2J3P2E3 HK3X4G5S3X4 ZO8C7C4A10P5文档编码:CG5K2J3P2E3 HK3X4G5S3X4 ZO8C7C4A10P5文档编码:CG5K2J3P2E3 HK3X4G5S3X4 ZO8C7C4A10P5文档编码:CG5K2J3P2E3 HK3X4G5S3X4 ZO8C7C4A10P5文档编码:CG5K2J3P2E3 HK3X4G5S3X4 ZO8C7C4A10P5文档编码:CG5K2J3P2E3 HK3X4G5S3X4 ZO8C7C4A10P5文档编码:CG5K2J3P2E3 HK3X4G5S3X4 ZO8C7C4A10P5文档编码:CG5K2J3P2E3 HK3X4G5S3X4 ZO8C7C4A10P5文档编码:CG5K2J3P2E3 HK3X4G5S3X4 ZO8C7C4A10P5文档编码:CG5K2J3P2E3 HK3X4G5S3X4 ZO8C7C4A10P5文档编码:CG5K2J3P2E3 HK3X4G5S3X4 ZO8C7C4A10P5文档编码:CG5K2J3P2E3 HK3X4G5S3X4 ZO8C7C4A10P5文档编码:CG5K2J3P2E3 HK3X4G5S3X4 ZO8C7C4A10P5文档编码:CG5K2J3P2E3 HK3X4G5S3X4 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ZO8C7C4A10P5文档编码:CG5K2J3P2E3 HK3X4G5S3X4 ZO8C7C4A10P5文档编码:CG5K2J3P2E3 HK3X4G5S3X4 ZO8C7C4A10P5文档编码:CG5K2J3P2E3 HK3X4G5S3X4 ZO8C7C4A10P5文档编码:CG5K2J3P2E3 HK3X4G5S3X4 ZO8C7C4A10P5文档编码:CG5K2J3P2E3 HK3X4G5S3X4 ZO8C7C4A10P5文档编码:CG5K2J3P2E3 HK3X4G5S3X4 ZO8C7C4A10P5文档编码:CG5K2J3P2E3 HK3X4G5S3X4 ZO8C7C4A10P5文档编码:CG5K2J3P2E3 HK3X4G5S3X4 ZO8C7C4A10P5文档编码:CG5K2J3P2E3 HK3X4G5S3X4 ZO8C7C4A10P5文档编码:CG5K2J3P2E3 HK3X4G5S3X4 ZO8C7C4A10P5精品资料欢迎下载故221121132222132111111i1111111()11211i1111zzzzwwwwwzwwzww.18.求出将割去负实轴Re()0z,Im(z)=0 的带形区域Im()22z映射为半带形区域 Im()w,Re(w)0 的映射.解：用1ezw将区域映为有割痕(0,1)的右半平面Re(w1)0；再用1211ln1www将半平面映为有割痕(-,-1的单位圆外域；又用32iww将区域映为去上半单位圆内部的上半平面；再用43lnww将区域映为半带形0Im(w4)0；最后用42iww映为所求区域，故e1lne1zzw.19.求将 Im(z)1 去掉单位圆|z|0 的映射.解：略.20.映射coswz将半带形区域0Re(z)0 保形映射为平面上的什么区域.解：因为1cos()2izizwzee可以分解为w1=iz，12eww，32211()2www由于coswz在所给区域单叶解析，所以（1）w1=iz将半带域旋转2，映为 0Im(w1),Re(w1)0.（2）12eww将区域映为单位圆的上半圆内部|w2|0.（3）2211()2www将区域映为下半平面Im(w)0.-第 7 页，共 7 页精品p d f 资料可编辑资料-文档编码:CG5K2J3P2E3 HK3X4G5S3X4 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