1.3 解方程组 ) 2(z)_(1)-(z)_(2)=i (1+i)(z)_(1)+i(z)_(2)=4-3i .

设z=(（1+i）（3-i）)/(（1+2i）（sqrt(2)+i）)，则z的模为（ ）A. $\frac{2}{\sqrt{3}}$B. $\frac{2\s

[单选题]复数z满足(1+i)z=1-i，则|Z|=()。A.√2/2B.√2C.2D.1

5.4已知复势为:(1) W(z)=(1+i)z ;(2) (z)=(1+i)ln (dfrac (z+1)(z-4)); (3) W(z)=-6iz+-|||

[单选题]若复数z1=1+i，z2=3-i，则z1·z2=（ ）A．4+2 i B. 2+ i C. 2+2 i D.3

设(z)=1-overline (z), _(1)=2+3i _(2)=5-i, 则 ([ f({z)_(1)-(z)_(2))-|||-__等于设等于

z=1+i，那么z=1+iz=1+iz=1+i，那么

下列命题错误的是：((z)_(1)z)=(l)_(n)(z)_(1)+(I)_(n)(z)_(2)((z)_(1)z)=(l)_(n)(z)_(1)+(I)_(

1.解方程z^2-4iz-(4-9i)=0.1.解方程$z^{2}-4iz-(4-9i)=0$.

1.设z=(1-sqrt(3)i)/(2)，求|z|及Arg z.1.设$z=\frac{1-\sqrt{3}i}{2}$，求|z|及Arg z.

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