(z)=(z)^2+dfrac (1)({z)^2+1}，则其解析区域为( )(z)=(z)^2+dfrac (1)({z)^2+1}(z)=(z)^2+dfrac (1)({z)^2+1}(z)=(z)^2+dfrac (1)({z)^2+1}(z)=(z)^2+dfrac (1)({z)^2+1}

(z)=(z)^2+dfrac (1)({z)^2-1}，则其解析区域为（）(z)=(z)^2+dfrac (1)({z)^2-1}(z)=(z)^2+dfra

求下列函数的奇点:(1)dfrac (z+1)(z({z)^2+1)}; (2)dfrac (z+1)(z({z)^2+1)}求下列函数的奇点:(1)

[单选题](z)=(z)^2+dfrac (1)({z)^2-1}f(z)=( )[单选题]A.B.C.D.

[题目]如果复数z1,z2,z3满足等式 dfrac (({z)_(2)-(z)_(1))}(({z)_(3)-(z)_(1))}=dfrac (({z)_(1

1.下列函数有些什么奇点?如果是极点,指出它的级数:-|||-(1) dfrac (1)(z{({z)^2+1)}^2} ;-|||-(2) dfrac (si

5.7 求出下列函数在有限孤立奇点处的留数:-|||-(1) dfrac ({e)^x-1}(z);-|||-(2) dfrac ({z)^7}((z-2){(

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

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(z)=dfrac (z+1)(z({z)^2+1)} 在复平面上只有-|||-8.[判断题]函数-|||-一个奇点 z=0 。() ()-|||-A 对-||

[ dfrac (sin z)({z)^2},0] =-|||-A 1-|||-B .-1-|||-C dfrac (1)(2)