e的z次方的模

将数字提出来(2^20)*(z^20)=(3^10)*(z^10)所以z^10=(3^10)/(4^10)因为是偶次幂,z=3/4或-3/4

（1+z+z^2/2!+...+z^n/n!+o(z^n))/(1-z)展开式应该就是这样吧,看你要保留到几项了.视你的具体情况而定.再问：答案是1+z+z2次方+z3次方…………再答：那这样不对啊(

(太麻烦拉,给点分啊!)设v=x*x-y*y,u=exp{xy}那么dv/dx=2x(这里应该用偏导符号,代替一下),dv/dy=2y,du/dx=y*exp{xy},du/dy=x*exp{xy}那

e^z-xyz=0对x求导əz/əx=(z'x)e^z-yz-xy(z'x)//z'x表示z对x的导数,下同对y求导əz/əy=(z'y)e^z-xz-xy(z

z'x=2e^(2x+y)z'y=e^(2x+y)所以dz=2e^(2x+y)dx+e^(2x+y)dy

已知二元函数z=f[x²-y²,e^(xy)]求∂²z/∂x∂y设z=f(u,v),u=x²-y²,v=e^(xy

(y^2+2xy-cos(y+z))/(e^z+cos(y+z))再问：没有过程吗？再答：求导：e^z*dz-y^2-2xy+cos(y+z)(1+dz)=0把含有dz的项移到一起：(e^z+cos(

求二元函数全微分z=f[x²-y²,e^(xy)]设z=f(u,v),u=x²-y²,v=e^(xy)则dz=(∂f/∂u)du+(&#

你想说这个问题?z=e^(x^2+2xy)应该是y=e^(x^2+2xy)(2x+2y）i+e^(x^2+2xy)2xj

z=arctan(x*e^x)z'={1/[1+(x*e^x)^2]}*(x*e^x)'(x*e^x)'=x'*e^x+x*(e^x)'=e^x+x*e^x=(x+1)*e^x所以dz/dx=(x+1

e^z=xyz两边对x求偏导e^z*z'(x)=y(z+x*z'(x))z'(x)=yz/(e^z-xy)∂z/∂x=yz/(e^z-xy)原式对y求偏导e^z*z'(y)=x

1e^z=xyze^zz'x=yz+xyz'xz'x=yz/(xy-e^z)=yz/(xy-xyz)=z/(x-xz)类似z'y=z/(y-yz)dz=[z/(x-xz)]dx+[z/(y-yz)]d

exp(exp(1))

∵z的n次方=1,∴z的（n+1）次方=z.又∵1+z.+z的n次方为等比数列前n+1项和,公比为z,当z≠1时,根据等比数列求和公式,得1+z.+z的n次方=（1-（z的（n+1）次方））/（1-z

设z=x+iyf(z)=e^z=e^(x+iy)=e^x·e^(iy)=e^xcosy+ie^xsinyRe[f(z)]=e^xcosy,Im[f(z)]=e^xsiny令u(x,y)=e^xcosy

求二元函数全微分z=f[x²-y²,e^(xy)]设z=f(u,v),u=x²-y²,v=e^(xy)则dz=(∂f/∂u)du+(&#

x^4＋y^4＋z^4－2x^2y^2－2x^2z^2－2y^2z^2＝(x^4＋y^4－2x^2y^2)＋z^4－2x^2z^2－2y^2z^2＝(x^2－y^2)^2－2(x^2－y^2)z^2＋

e^z=1+√3i=2e^i(π/3)=e^[ln2+i(2kπ+π/3)]得：z=ln2+i(2kπ+π/3),这里k为任意整数

e^(-xy)-x^2*y+e^z=z,令F(x,y,z)=e^(-xy)-x^2*y+e^z-z=0分别对F取x,y,z的偏导数,可得əF/əx=e^(-xy)*(-y)-2xy

若z是实数的话,则z=ln(1+√3)若z是复数,则∵exp(2πi)=1∴expz是周期函数,周期是2πi∴z=ln(1+√3)+2kπi,(k∈Z)也是解∴解为z=ln(1+√3)+2kπi