1 (z×(z-1)×(z 2)²)在圆周z=3 2处的积分,利用留数
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|f(z1+z2)|=|f(2+3i+2+i)|=|f(4+4i)|=|(|1-4-4i|)|=|(|-3-4i|)|=|√(3²+4²)|=5
复数Zn=[(1-i)/2]^n.(n=1,2,3,...).∴|Z(n+1)-Zn|=|[(1-i)/2]^n|×|[(1-i)/2]-1|=|(1-i)/2|^n×|(1+i)/2|=[(√2)/
设z=x+yi(x,y∈R),∵(1+3i)z=(1+3i)(x+yi)=(x-3y)+(3x+y)i∈R∴虚部3x+y=0,即y=-3x &
设z=x+yi,(x、y∈R),则(1+3i)•z=(x-3y)+(3x+y)i为纯虚数,∴x-3y=0,3x+y≠0,∵|ω|=|z2+i|=52,∴|z|=x2+y2=510;又x=3y.解得x=
z1=1-2i,1/z1=1/(1-2i)=(1+2i)/5z2=3+4i,1/z2=1/(3+4i)=(3-4i)/251/z=1/z1+1/z2=(1+2i)/5+(3-4i)/25=(5+10i
√5i,z2=1-√3i,z=4z1^4/3z2
1/z=(z1+z2)/(z1z2)z=(5+10i)(3-4i)/(5+10i+3-4i)=(15+40-20i+30i)/(8+6i)=(55-10i)(8-6i)/(8+6i)(8-6i)=5(
1/(1-2i)+1/(3+4i)=(1+2i)/5+(3-4i)/25=(8+6i)/25所以z=25/(8+6i)=25(8-6i)/100=2-(3/2)i
1)若Z⒉+2|Z|=a(a≥0)求复数Z|Z|和a都是实数,所以Z是实数,Z=a/42)若复数Z满足|Z|=|Z+2+2i|,则|Z-1+2i|的最小值是________|Z-0|=|Z-(-2-2
设z2=x+yiz1*z2=(1+3i)(x+yi)=x-3y+(3x+y)i+为纯虚数,则x=3yz2=3y+yi|z2|=y√10|(z+2i)|=2√2|z2/(z+2i)|=y√10/(2√2
z1=1+2i,z2=2-i,z1+z2=1+2i+2-i=3+i1/z=3+iz=1/(3+i)=(3-i)/(3+i)(3-i)=1/10(3-i)=3/10-1/10i
1/z=1/(5+10i)+1/(3-4i)=(3-4i+5+10i)/(5+10i)(3-4i)=(8+6i)/(15-20i+30i+40)=(8+6i)/(55+10i)z=(55+10i)/(
1/z1=1/(5+10i)=1/[5(1+2i)]=(1/5)×[(1-2i)]/[(1+2i)(1-2i)]=(1-2i)/251/z2=1/(3-4i)=(3+4i)/[(3-4i)(3+4i)
x/(y+z)+y/(z+x)+z/(x+y)=1所以x/(y+z)=1-[y/(z+x)+z/(x+y)]y/(z+x)=1-[x/(y+z)+z/(x+y)]z/(x+y)=1-[x/(y+z)+
等于0.x/(y+z)=1-[y/(z+x)+z/(x+y)]y/(z+x)=1-[x/(y+z)+z/(x+y)]z/(x+y)=1-[x/(y+z)+y/(z+x)]x2/(y+z)+y2/(z+
1z=1z1+1z2=z1+z2z1z2∴z=z1z2z1+z1又∵z1=5+10i,z2=3-4i∴z=(5+10i)(3−4i)5+10i+3−4i=55+10i8+6i=(55+10i)(8−6
∵|z|=1,∴z=cosθ+isinθ,∴|2z2-z+1|=|2(cosθ+isinθ)2-(cosθ+isinθ)+1|=|(2cos2θ-cosθ+1)+(2sin2θ-sinθ)i|=(2c
设z1,z2是一个实系数一元二次方程的两个虚根,则z1和z2是互为共轭的虚数,可分别设为a+bi,a-bi,由z1^2=z2,可得z1^2=a^2-b^2+2abi,z2=a-bi故有:a^2-b^2