Z的模长=1,Z² 2Z Z分之1
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由Z-2的模等于2可知|Z-2|=2得Z=0或Z=4因为Z+Z分之1属于R所以(Z+1)/Z属于R所以Z=0舍去所以Z=4
X:Y:Z=1:2:3因为:14(XX+YY+ZZ)=(X+2Y+3Z)^214(XX+YY+ZZ)-(X+2Y+3Z)^2=013X^2+10Y^2+5Z^2-4XY-6XZ-12YZ=0(4X^2
1、析:设z=a+bi,b≠0则z+1/z=a+bi+1/(a+bi)=a+bi+(a-bi)/(a^2+b^2)=(a+a/(a^2+b^2)+[b-b/(a^2+b^2]i,∴b-b/(a^2+b
设z=x1+y1*i,z'=x2+y2*i,z+2z’为纯虚数得x1=-2x2代入:10z^2+5z’^2=2zz’得:49x^2-10(y1)^2-5(y2)^2+2y1*y2=0,-42x2*y1
则由题意得,(z+1)/z=2(cosπ/3+sinπ/3*i),设z=a+bi(a+bi+1)/a+bi=2(cosπ/3+sinπ/3*i)a+1+bi=(a-sqrt(3))+(sqrt(3)a
虚数z满足|z|=1,z²+2z+1/z
(X+Y+Z)*(X+Y+Z)=XX+YY+ZZ+2(XY+YZ+XZ)=1,又XY+YZ+XZ=0,所以XX+YY+ZZ=1
X=3K,Y=K,Z=2K(2X方-2Y方+5Z方)/(XY+YZ+ZX)=(2*9-2+5*4)/(3+2+6)=36/11
|z|=1且z≠±i,则可设z=cosθ+isinθz/(1+z²)=(cosθ+isinθ)/[1+(cosθ+isinθ)²]=(cosθ+isinθ)/(1+cos²
设z=x+yiz+1/z=(x+yi)+1/(x+yi)=(x+yi)+(x-yi)/(x²+y²)=x+x/(x²+y²)+[y-y/(x²+y&s
解题思路:先进行复数的乘除运算,把具体的复数的值代入,整理成最简形式,得到复数相等的条件,使得复数的实部和虚部分别相等,得到关于a和b的方程组,解方程组即可.解题过程:
x2+y2-z2+2xy/x2-y2+z2-2xz=(x+y)2-z2/(x-z)2-y2=(x+y-z)(x+y+z)/(x-y-z)(x-z+y)=(x+y+z)/(x-y-z)然后就是代入了
因为模[(z+1)/z]=2arg[(z+1)/z]=π/3所以(z+1)/z=2(cosπ/3+isinπ/3)1+1/z=1+√3i1/z=√3iz=1/[√3i]=-√3/3i
设Z=a+bia×a+b×b=1①(a+1)×(a+1)+b×b=1②连立①②得:a=-1/2,b=-(根号3)/2则Z-1=-3/2-(根号3)/2Z-1的模=根号3
设z=x+yi,则z+z-+zz-=0x+yi+x-yi+x^2+y^2=0x^2+y^2+2x=0(x+1)^2+y^2=1所以复数z的轨迹是以(-1,0)为圆心,以1为半径的圆
a=1;z=1+iz+1/z=1+1/z=1+1/1-z=1+z/2+1=3/2+1/2z再问:可以明白一点不〜谢了!
|z|=√(1+1)=√2z=√2(√2/2+i√2/2)=√2(cos45°+isin45°)所以辐角主值aryZ=45°
(x+y)^2=(x-y)^2+4xy=64+4(-z^2-16)=-4z^2=0所以(x+y)^2=0所以x+y=0x-y=8x=4,y=-4z=0
可设z=a+bi.(a,b∈R).由题设可知,z+z拔=(a+bi)+(a-bi)=4.===>a=2.===>z=2+bi.∴z-z拔=(2+bi)-(2-bi)=2bi.由题设|2bi|×|1+i