已知z=a-i 1-i
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∵a+i1−i=(a+i)(1+i)(1−i)(1+i)=a−1+(a+1)i2为纯虚数,∴a−12=0且a+12≠0,解得:a=1.故选:A.
∵复数z=i1−i=i•(1+i)(1−i)•(1+i)=−12+12i故复数z=i1−i(i为虚数单位)的虚部是12故答案为:12
复数z=i1+i=i(1−i)(1+i)(1−i)=1+i2=12+i2,∴复数z的模为14+14=22,故选A.
∵复数z=1+i1−i=(1+i)2(1−i)(1+i)=2i2=i,∴.z=-i,故答案为:-i.
∵z=a-i1-2i=(a-i)(1+2i)(1-2i)(1+2i)=(a+2)+(2a-1)i5∵z为实数∴2a-1=0解得a=12故答案为:12.
z=x+3i1−i(x∈R)=(x+3i)(1+i)2=(x−3)+(x+3)i2,∵复数z=x+3i1−i(x∈R)是实数,∴x+3=0,∴x=-3.故选A.
∵复数z=2−2i1+i=(2−2i)(1−i)(1+i)(1−i)=−4i2=-2i,∴复数z的共轭复数等于2i故选A.
由z=2i1+i=2i(1−i)(1+i)(1−i)=2+2i2=1+i.所以z2=(1+i)2=1+2i+i2=2i.故选A.
i=i1+i2=2sin(wt+30º)+4sin(wt-45º)=2(sinwtcos30º+sin30coswt)+4(sinwtcos45º-sin45&
∵复数z=a+i1−i=(a+i)(1+i)(1−i)(1+i)=a−1+(1+a)i2是纯虚数,∴a−12=0a+12≠0,解得a=1.故答案为:1.
复数z=2i1+i=2i(1−i)(1+i)(1−i)=1+i所以它的共轭复数为:1-i故选A
由Z=1−i1+i=(1−i)(1−i)(1+i)(1−i)=−2i2=−i,所以1+Z+Z2+Z3+Z4=1-i+(-i)2+(-i)3+(-i)4=1-i-1+i+1=1.故答案为1.
复数z=1+3i1−i=(1+3i)(1+i)(1−i)(1+i)=−2+4i2=-1+2i,则z的实部为-1.故选:D.
设z=x+yi,则x²+y²=25(1),又(3+4i)(x+yi)=3x-4y+(3y+4x)i为纯虚数,所以3x-4y=0(2)4x+3y≠0(3)由(1)(2),解得x=4,
复数a+2i1−i=(a+2i)(1+i)(1−i)(1+i)=a−2+(a+2)i2又复数a+2i1−i在复平面内所对应的点在虚轴上所以a-2=0,即a=2故答案为2
z=a+3i1−2i=(a+3i)(1+2i)(1−2i)(1+2i)=a−6+(2a+3)i5它是纯虚数,所以a=6,|a+2i|=|6+2i|=36+4=210故答案为:210
复数z=2−4i1+i=(2−4i)(1−i)(1+i)(1−i)=−2−6i2=-1-3i,故z的共轭复数等于-1+3i,故选C.
设z=x+yi,代入方程z•.z+2i•z=4+2i,得x2+y2+2xi-2y=4+2i故有x2+y2−2y=42x=2解得x=1y=−1或3故 z=1-i或z=1+3i故答案为:1-i或
3+i1+i=(3+i)(1−i)(1+i)(1−i)=3+i−3i−i22=4−2i2=2−i.故答案为:2-i.
∵Z=1-3i1+i=(1-3i)(1-i)(1+i)(1-i)=-2-4i2=-1-2i∴复数的实部是-1,故选B.