(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
3.求下列函数在给定点的Taylor级数,并指出收敛半径.-|||-(1) dfrac (z-1)(z+1) , _(0)=1 :-|||-(2)sinz^2,
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){(
(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+dfr
z=0是函数(z)=dfrac (z)(sin {z)^2cdot ((e)^z-1)}的几级极点A 1 B 2 C 3 D 4z=0是函数的
曲线C为正向圆周|z-1|=3, (int )_(c)^3dfrac (3)(2)dz=|z-1|=3, (int )_(c)^3dfrac (3)(2)dz=
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
1.下列函数有些什么奇点?如果是极点,指出它的级:(1)(1)/(z(z^2)+1)^(2); (2)(sin z)/(z^3); (3)(1)/(z^3)
[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
5.利用留数计算下列积分.-|||-(3) (int )_(|z|=2)dfrac ({e)^2z}((z+1){(z-1))^2}dz