若((dfrac {a+i)(1-i))}^2=-1，则a=；A 0 B 2 C 1 D 3

A . 0 B . =-dfrac (1-5i)(2+3i) C . =-dfrac (1-5i)(2+3i)D . =-dfrac (1-5i)(2+3i)设

求复数=dfrac (2i)(1-i)+dfrac (2(1-i))(i)的模与主辐角。求复数的模与主辐角。

dfrac (1)(z)-dfrac (3i)(1-i)的共轭复数为 dfrac (1)(z)-dfrac (3i)(1-i) D.以上都不对的共轭复数为D.以

[问答题]若复数z=a/(1-i)+(1-i)/2是纯虚数，则实数a等于________。

(D) =dfrac ({mu )_(0)(I)_(1)(I)_(2)}(2pi )ln dfrac (b)(a)-|||-Ai-|||-I1-|||-C I2

1.9 利用复数的三角表示计算下列各式:-|||-(1) (1+i)(1-i) ;-|||-(2) (-2+3i)/(3+2i) ;-|||-(3) ((dfr

已知复数=dfrac (2-2i)(1+sqrt {3)i}，则=dfrac (2-2i)(1+sqrt {3)i}=dfrac (2-2i)(1+sqrt {

4.设 (z)=dfrac (1)(z)-zsin dfrac (1)({z)^2}, 则 [ f(z),0] =-|||-(A)1; (B)2; (C)0;

1.求下列复数z的实部与虚部、共轭复数、模与辐角:-|||-(1) 1/(3+2i);-|||-(2) dfrac (1)(i)-dfrac (3i)(1-i)

3、设 (x,y)=arctan dfrac (x)(y), 则 (1,1)=-|||-(A)1; (B)0; (C) dfrac {1)(2),dfrac