i为虚数单位,若(根号3 i)z=根号3-1
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(1-z)/(1+z)=(-1+i)/(3+i)(1-z)(3+i)=(-1+i)(1+z)3+i-3z-zi=-1-z+i+zi2z+2zi=42z(1+i)=4z=2/(1+i)=2(1-i)/(
Z=(3-i)/i=(3-i)i/i*i=-(3-i)i=-1-3iZ的绝对值=根号(1^2+3^2)=根号10
|Z+(1-i/1+i)^6|≤|-根号3i||Z+(-i)^6|≤根号3|Z-1|≤根号3Z对应半径为根号3的园面SZ=3π
z=(1-√3i)/(√3-i)=(1-√3i)(√3+i)/(√3+i)(√3-i)=(√3+i-3i+√3)/(3+1)=(2√3-2i)/4=√3/2-(1/2)i所以虚部是-1/2
方法一:z(1+√3i)=1+i所以z=(1+i)/(1+√3i)=(1+i)(1-√3i)/[(1+√3i)(1-√3i)]=[(1+√3)+(1-√3)i]/4|z|=√[(1+√3)²
z1*z2=6+3i-2i-i^2=7+i=x+yi所以x=7y=1
(1+i)z=2i³z=2i³/(1+i)=-2i(1-i)/(1+i)(1-i)=(-2i-2)/(1+i)=-1-i
(4-3i)z=5i然后左右同时乘以4+3i得到25z=5i(4+3i)得到z=-3/5+4i/5
为了输入方便,将z^-用大写Z表示则z+Z=√6,(z-Z)*i=-√2设z=x+yi,则Z=x-yi∴2x=√6,即x=√6/22yi*i=-√2即2y=√2即y=√2/2(1)z=(√6/2)+(
∵(z+i)i=-3+i,∴-i•i(z+i)=-i(-3+i),化为z+i=3i+1,∴z=2i+1,故选:A.
设Z=a+biZ的共轭复数为a-bi所以由题2a=√6a=√6/22bi*i=-√2-2b=-√2b=√2/2所以Z=√6/2+√2/2i
z=(2+3i)(-3-2i)/(-3+2i)(-3-2i)=(-6-4i-9i+6)/(9+4)=-13i/13=-i所以虚部是-1
z=(2i+1/i)^3=(2i^2+1)^3/i^3=(-2+1)^3/(-i)=1/i=-i│z-1│=│-i-1│=根号2
z=(3+4i)/(1+2i)=(3+4i)(1-2i)/(1+2i)(1-2i)=(3-6i+4i+8)/5=(11-2i)/5∴|z|=√[(11/5)²+(2/5)²]=√5
∵f(z+i)=z-3i,设t=z+i,则z=t-i,∴f(t)=t-i-3i=t-4i,∴f(2i)=2i-4i=-2i,∴|f(2i)+1|=|1-2i|=5,故答案为:5.
z=(根号3-i)/[1+i根号3]=(根号3-i)*[1-i根号3]/{[1-i根号3][1+i根号3]}=(-4i)/(1+3)=-iz的模为1
(1+√3i)z=iz=i/(1+√3i)=i(1-√3i)/(1²-3i²)=(i+√3)/42.2/(1-i)=2(1+i)/(1²-i²)=1+i若1+i
已知复数z=√3-i(i为虚数单位)则4/z=4/(√3-i)=4(√3+i)/(√3-i)(√3+i)=4(√3+i)/(3+1)=√3+i;您好,很高兴为您解答,skyhunter002为您答疑解
(1)(Z+i)i=3-1;Z*i-1=3-1;Z*i=3;两边同乘i;变为-Z=3i;Z=-3i为所求.Z的模:(0的平方+(-3)的平方)开根号=3(2)Z=-3i,a=0,b=-3,A点坐标(0
设z=a+bi∴i(a+bi+i)=ai-b-1=-(b+1)+ai=-3+2i根据对应关系b+1=3,a=2∴a=2,b=2