(z)=(3+dfrac ({z)^2}(2))sin z在z=0处的展开式中(z)=(3+dfrac ({z)^2}(2))sin z的(z)=(3+dfrac ({z)^2}(2))sin z则A=

sin (x+2y-3z)=x+2y-3z, 则 dfrac (partial z)(partial x)+dfrac (partial z)(partial

[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

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

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

(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

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.设 sin (x+2y-3z)=x+2y-3z, 证明: dfrac (partial z)(partial x)+dfrac (partial z)(pa

1.已知 sin (3x-2y+z)=3x-2y+z, 则 dfrac ({partial )^2z}(partial xpartial y)= __

已知复数(z)=(z)^2+3z-2，则(z)=(z)^2+3z-2分别为(z)=(z)^2+3z-2(z)=(z)^2+3z-2(z)=(z)^2+3z-2(