设a,b∈c,且ab+2ai-2bi+1=0,若|b|=0,求复数a
已知复数z=a+bi,(a>0,b∈R)若z^2=b+ai,则z=?
若a,b∈R,(5+bi)+(b-3i)-(2+ai)=0,那么复数a+bi的模为?
若a-2i=bi+1(a、b∈R),复数z=b+ai,则z.z
已知复数z=a+bi(a,b∈R且ab≠0),且z(1-2i)为实数,则ab=( )
设a.b为实数,若复数1+2i/a+bi=1+i,求a.b
设a.b为实数,若复数1+2i/a+bi=1+i,则 a+bi =
设复数z=a+bi(a,b∈R,b>0),z^2/(1+z)和z/(1+z^2)均为实数.求z
设a,b为实数,若复数1+2i/a+bi=1+i
已知复数z1=a+bi,z2=-b+ai(a,b∈R,且ab≠0),则在复平面内,z1,z2对应的点与原点组成的三角形是
设z=a+bi,且a,b满足a(1+i)³+(2-5i)=bi-4,则z的共轭复数=
若复数2-ai=i(1-bi),则a+bi
设a,b为实数,若复数(1+i)(a+bi)=1+2i,则a= b=