已知2z+|z|=2+6i，求复数z．

设方程\(z^2+y^2+z^2 = 4z\)确定函数\(z=z(x,y)\)，则\( { { {\partial ^2}z} \over {\partial {x^2}}} =\) A: \( { { { { (2 - z)}^2} + {x^2}} \over { { {(2+ z)}^3}}}\) B: \( { { { { (2 - z)}^2} + {x^2}} \over { { {(2 - z)}^3}}}\) C: \( { { { { (2 - z)}^2} -{x^2}} \over { { {(2 - z)}^3}}}\) D: \( { { { { (2 + z)}^2} + {x^2}} \over { { {(2 - z)}^3}}}\)

9. 已知函数$z=z(x,y)$由${{z}^{3}}-3xyz={{a}^{3}}$确定，则$\frac{{{\partial }^{2}}z}{\partial x\partial y}=$（ ） A: $\frac{z({{z}^{4}}-2xy{{z}^{2}}-{{x}^{2}}{{y}^{2}})}{{{({{z}^{2}}-xy)}^{3}}}$ B: $\frac{z({{z}^{4}}-2xy{{z}^{2}}-xy)}{{{({{z}^{2}}-xy)}^{2}}}$ C: $\frac{z({{z}^{3}}-2xyz-{{x}^{2}}{{y}^{2}})}{{{({{z}^{2}}-xy)}^{3}}}$ D: $\frac{z({{z}^{3}}-2xy{{z}^{2}}-{{x}^{2}}y)}{{{({{z}^{2}}-xy)}^{3}}}$

1）z^2=z拔（2）z^2+|z|=0

【简答题】设 z 1 =4 + 3i , z 2 =2 - 3i ,计算 z 1 · z 2

\( xoz \) 坐标面上的直线\( x = z - 2 \)绕\( z \)轴旋转而成的圆锥面的方程为（ ） A: \( {x^2} - {y^2} = {(z - 2)^2} \) B: \( {x^2} + {y^2} = {(z - 2)^2} \) C: \( {z^2} + {y^2} = {(x - 2)^2} \) D: \( {z^2} + {x^2} = {(y - 2)^2} \)

f(z)=e^(z^2)*sin(z^2),求f(z)展成Z的幂级数,

若复数z 1＝1＋i ，z 2＝3－i ，则z 1·z 2等于

z=0为f(z)=z^2 (e^(z^2 )-1)的 级零点,

若复数z满足：z+.z=2，z?.z=2，则|z-.z|=______．