3.7 计算 (theta )_(1)=1=3=dfrac (1)((z-i)(z+2))dz .

(3.)求下列极限:-|||-(1) lim _(xarrow infty )dfrac (1)(1+{z)^2} ;-|||-(3) lim _(zarrow

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

5.利用留数计算下列积分.-|||-(3) (int )_(|z|=2)dfrac ({e)^2z}((z+1){(z-1))^2}dz

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=

12.计算下列各积分，C为正向圆周：1）oint(z^15)/((z^2)+1)^(2(z^4+2)^3)dz，C：|z|=3；2）oint(z^3)/(1+z

[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

3.6 计算 (int )_(c)dfrac (1)({z)^2-z}dz, 其中C为圆周 |z|=2.

设C：|z-2|=5为正向圆周，则int dfrac (2{z)^3+3(z)^2+2z+1}(z)dz=()A、2πіB、πі; C、i;D、0;设C：|z-

曲线C为正向圆周|z|=2, (int )_(c)dfrac (cos z)({(z-1))^3}dz=曲线C为正向圆周A.0B.C.D.

设 C 为正向圆周 |z+1|=2，n 为正整数，则积分 oint_(C) (dz)/((z-i)^n+1) 等于（ ） 设 $C$ 为正向圆周 $|z+1|=