若 z^2 = (overline(z) )^2，则必有()

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

[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

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

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、

已知复数(z)=(z)^2+3z-2，则(z)=(z)^2+3z-2分别为(z)=(z)^2+3z-2(z)=(z)^2+3z-2(z)=(z)^2+3z-2(

7.若f(z)=xy^2+ix^2y，则f(z)（）A. 仅在直线y=x上可导B. 仅在直线y=-x上可导C. 仅在(0,0)点解析D. 仅在(0,0)点可导

(z)=(z)^2+dfrac (1)({z)^2-1}，则其解析区域为（）(z)=(z)^2+dfrac (1)({z)^2-1}(z)=(z)^2+dfra

(z)=(z)^2+dfrac (1)({z)^2+1}，则其解析区域为( )(z)=(z)^2+dfrac (1)({z)^2+1}(z)=(z)^2+dfr

1.[单选题]若f(z)=x^2+iy^2，则f(z)（）A. 在全平面上解析B. 仅在直线y=x上可导C. 仅在直线y=-x上可导D. 仅在(0,0)点可导

(z)=(3+dfrac ({z)^2}(2))sin z在z=0处的展开式中(z)=(3+dfrac ({z)^2}(2))sin z的(z)=(3+dfra