F(2z)

F(Z)=Z*Z-Z-2分之一如何展成Z的幂级数,并求出收敛半径.F(Z)=(Z*Z-Z-2)分之一F(z)=1/(z^2-z-2)=(1/3)(1/(z-2)-1/(z+1))1/(z-2)=(-1/2)/(1-z/2)=(-1/2)(1

已知f(z)=e^z/z^2,求Res(f(z),0)(Resf(0))f(z)=e^z/z^2=(1+z+z^2/2+...)/z^2=1/z^2+1/z+1/2+...,因此Resf(0)=1.Res(f(z),0)=1.解法：z=0,

已知f(z)=e^z/z^2,求Res(f(z),0)(Resf(0))e^z=1+z+z^2/2+...故f(z)=e^z/z^2=1/z^2+1/z+1/2+...1/z的系数为1故Res(f(z),0)=1

f'(z)=(z+1)'(2-z)+(z+1)(2-z)'如何计算(z+1)'=z'+1'=1+0=1(2-z)'=2'-z'=0-1=-1所以,f'(z)=(z+1)'(2-z)+(z+1)(2-z)'=（2-z）-（z+1）=-2z+1

f(z)=z^2/{(z^2+1)*(z^2+9)}求Res(z=i)f(z)和Res(z=3i)f(z)(积分区间为0到正无穷=1/2∫f(x)dx(积分区间为负无穷到正无穷）=ipai{Res[f(z),i]+Res[f(z),3i]}

f(z）=2z+z'-3if(z'+i)=6-3i,则f(-z)=?z',z是复数!因为f(z）=2z+z'-3i,把z'+i代入有：f(z'+i）=2(z'+i)+z'-3i=3z'-i又因为：f(z'+i）=6-3i.令z'=x+yi.

复变函数f(z)=z^2/{(z^2+1)*(z^2+9)}求Res(z=i)f(z)和Res(z=3i)f(z)见图

复变函数f(z)=z^2/{(z^2+1)*(z^2+9)}求Res(z=i)f(z)和Res(z=3i)f(z)当z=i时,代入式子得f（z）=i^2/{(i^2+1)(i^2+9)}=-1/8Res(f(z))=-1/8当z=3i时,代

将函数f(Z)=Z/Z+2展开成Z-2的幂级数f(z)=1-2/(z+2)=1-2/[(z-2)+5]=1-0.4*1/[1+(z-2)/5]=1-0.4*Σ【-（z-2)/5】^n(0到+∞）

F(Z)=1/(Z-1)(z-2)在Z=1处的泰勒展开式F(Z)=1/(Z-1)(z-2)=1/(z-2)-1/(z-1)第二项-1/(z-1)不必继续展开,只考虑第一项1/(z-2)=1/(z-1-1),令x=z-1,则第一项变为1/(x

求f(z)=z/(z+2)展开为z的泰勒级数...f(z)=1-2/(z+2)=1-1/[1+(z/2)]=1-1/[1-(-z/2)],根据1/(1-z)=1+z+z^2+...,所以f(z)=z/2-z^2/2^2+z^3/2^3-..

求积分计算f{|z|=pi}(z/(z+1))*(e^(2/(z+1)))dzf(z)=z/(z+1)*e^[2/(z+1)]设I=∫(|z|=π)f(z)dz因为在区域|z|而f(z)=[1-1/(z+1)]{1+2/(z+1)+(1/2

f(Z)=1/z(z+1)(z+4)在2http://hi.baidu.com/522597089/album/item/ee9535deeee8df9a74c63876.html#.....为了方便记3^x1=m>13^x2=n>1所以不

将函数f(z)=1/[(z-1)(z-2)]在|z|f(z)=1/(1-z)-1/(2-z)=∑z^n-1/2*1/(1-z/2)=∑z^n-1/2*∑(z/2)^n可以做吗?|z|可以做,对的

将函数f(z)=1/[(z-1)(z-2)]在|z|f(z)=1/(1-z)-1/(2-z)=∑z^n-1/2*1/(1-z/2)=∑z^n-1/2*∑(z/2)^n这里*表示乘号,∑表示n从0到无穷求和原式化成-1/(1-z)+1/[1+

（2）f(z)=z^2/(z^2+1)(z^2+9),求Resf(z)（z=i）和Resf(z)（z=3i）Res[f(z),i]={(z-i)*[z^2/(z^2+1)(z^2+9)]}[（z=i）下标]=i^2/(i+i)(i^2+9)

f(z)=z^2/(z^2+1)(z^2+9),求Resf(z)（z=i）和Resf(z)（z=3i）z=a+bi|z|=1+3i-z|z|=√(a^2+b^2)√(a^2+b^2)=1+3i-a-bi3-b=0b=3√(a^2+9)=1-

已知f(z+i)=z+2z-2i,则f(i)=?f(z+i)=z+2z-2i,则f(i)=?f(z+i)=z+2z-2i,令z=0,有：f(i)=-2i

已知f(z)=2z+z^2+(1+i),则f(i)的值是f(z)=(z+1)^2+i则f(i)=(i+1)^2+i=i^2+3i+1=3i

已知f(z)=2z+z²+(1+i)则f(i)的值是=2i-1+1+i=3i