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

等于
A. -4-4i
B. 4+4i
C. 4-4i
D. -4+4i

参考答案与解析:

相关试题

若复数z1=1£«i,z2=3£­i,则z1·z2=()

[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

  • 查看答案
  • 一、设f(z)=(1)/(2i)((z)/(overline(z))-(overline(z))/(z)),z≠0.试证:当z→0时,f(z)的极限不存在.

    一、设f(z)=(1)/(2i)((z)/(overline(z))-(overline(z))/(z)),z≠0.试证:当z→0时,f(z)的极限不存在.一、

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

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

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

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

  • 查看答案
  • 设z=((1+i)(3-i))/((1+2i)(sqrt(2)+i)),则z的模为( )

    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

  • 查看答案
  • 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)_(n)(z)_(2)((z)_(1)z)=

    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)_(

  • 查看答案
  • 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-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-

  • 查看答案
  • 5.设z_(1)及z_(2)是两复数.求证:(1)|z_(1)-z_(2)|^2=|z_(1)|^2+|z_(2)|^2-2mathrm(Re)(z_(1)overline(z_{2)});(2)|z

    5.设z_(1)及z_(2)是两复数.求证:(1)|z_(1)-z_(2)|^2=|z_(1)|^2+|z_(2)|^2-2mathrm(Re)(z_(1)ov

  • 查看答案
  • 1.设z=(1-sqrt(3)i)/(2),求|z|及Arg z.

    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.

  • 查看答案
  • 2.设f(z)=(e^z)/(z^2)-1,求Res(f(z),∞).

    2.设f(z)=(e^z)/(z^2)-1,求Res(f(z),∞).2.设$f(z)=\frac{e^{z}}{z^{2}-1}$,求Res(f(z),∞).

  • 查看答案