5.证明 (z)=cos (z+dfrac (1)(z)) 用z的幂表出的洛朗展开式中的系数为-|||-_(n)=dfrac (1)(2pi )(int )_(0)^2pi cos (2cos theta )cos ntheta dtheta =0 ,±1 ,±2 .....

参考答案与解析：

相关试题

证明:在(z)=cos (z+dfrac (1)(z)) 以z的各幂表出的洛朗展开式中的各系数为在(z)=cos (z+dfrac (1)(z)) 以z的各幂表出的洛朗展开式中的各系数为提示:令C为单

证明:在(z)=cos (z+dfrac (1)(z)) 以z的各幂表出的洛朗展开式中的各系数为在(z)=cos (z+dfrac (1)(z)) 以z的各幂表

  • 查看答案

    • 例8 已知 (z)=dfrac (1)(2pi )(|)_(|i|=1)dfrac (cos xi )({(xi -z))^3}ds, 证明:当 |z|neq 1 时,f(z)解析,-|||-并求f`

    例8 已知 (z)=dfrac (1)(2pi )(|)_(|i|=1)dfrac (cos xi )({(xi -z))^3}ds, 证明:当 |z|neq

  • 查看答案

    • 计算(int )_(c)_(c)dfrac (cos z)((z-dfrac {1)(2))(z-1)}dz，其中(int )_(c)_(c)dfrac (cos z)((z-dfrac {1)(2)

    计算(int )_(c)_(c)dfrac (cos z)((z-dfrac {1)(2))(z-1)}dz，其中(int )_(c)_(c)dfrac (co

  • 查看答案

    • 积分int dfrac (cos z)({(pi -z))^3}dz=______

    积分int dfrac (cos z)({(pi -z))^3}dz=______积分=______

  • 查看答案

    • 1.一平面电磁波能表示成 _(x)=0, _(y)=2cos [ 2pi times (10)^14(dfrac (z)(c)-t)+dfrac (pi )(2)] _(2)=0, 则表明:-|||

    1.一平面电磁波能表示成 _(x)=0, _(y)=2cos [ 2pi times (10)^14(dfrac (z)(c)-t)+dfrac (pi )(2

  • 查看答案

    • 把复数=((cos dfrac {2pi )(9)+isin dfrac (2pi )(9))}^3转化为三角形式A =((cos dfrac {2pi )(9)+isin dfrac (2pi )(

    把复数=((cos dfrac {2pi )(9)+isin dfrac (2pi )(9))}^3转化为三角形式A =((cos dfrac {2pi )(9

  • 查看答案

    • πm·s^(-1))=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)

    πm·s^(-1))=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T

  • 查看答案

    • 5.12 求下列各积分之值:-|||-(1) (int )_(0)^2pi dfrac (dtheta )(a+cos theta )(agt 1);-|||-(2) (int )_(0)^2pi d

    5.12 求下列各积分之值:-|||-(1) (int )_(0)^2pi dfrac (dtheta )(a+cos theta )(agt 1);-|||-

  • 查看答案

    • 7.设 +(z)^-1=2cos theta (zneq 0,theta 是Z的辐角),求证 ^n+(z)^-n=2cos ntheta .

    7.设 +(z)^-1=2cos theta (zneq 0,theta 是Z的辐角),求证 ^n+(z)^-n=2cos ntheta .

  • 查看答案

    • (单选题) (int )_(0)^2pi (x)^2cos xdx= ().-|||-

    (单选题) (int )_(0)^2pi (x)^2cos xdx= ().-|||-

  • 查看答案