设z=((1+i)(3-i))/((1+2i)(sqrt(2)+i)),则z的模为( )A. $\frac{2}{\sqrt{3}}$B. $\frac{2\s
设(z)=1-overline (z), _(1)=2+3i _(2)=5-i, 则 ([ f({z)_(1)-(z)_(2))-|||-__等于设等于
在下列复数中,使得^z=sqrt (3)+i成立的是( )^z=sqrt (3)+i ^z=sqrt (3)+i ^z=sqrt (3)+i ^
1.1.18 就以下各种情况,分别求arg z:-|||-(a) =dfrac (-2)(1+sqrt {3)i} ;-|||-(b) =dfrac (i)(-
1.设 ^2+(y)^2+(z)^2-z=0, 求 dfrac ({a)^2z}(a{y)^2}
2.设f(z)=(e^z)/(z^2)-1,求Res(f(z),∞).2.设$f(z)=\frac{e^{z}}{z^{2}-1}$,求Res(f(z),∞).
设 =X/3-Y/4,-|||-(1)求E(Z),D(Z);(2)求Cov(Y,Z).
3设 +2y+z-2sqrt (xyz)=0, 求 dfrac (partial z)(partial x) 及 dfrac (partial z)(parti
1.3 解方程组 ) 2(z)_(1)-(z)_(2)=i (1+i)(z)_(1)+i(z)_(2)=4-3i .
[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