当z=-1+3i时,|z+2-4i|有最小值,为√2.
9.导学号88184059已知复数z=("(-" 1+3i")(" 1"-" i")-(" 1+3i")" )/i,w=z+ai(a∈R),当|w/z|≤√2时,求a的取值范围.
解z=("(-" 1+3i")(" 1"-" i")-(" 1+3i")" )/i
=("(" 2+4i")-(" 1+3i")" )/i=(1+i)/i=1-i.
∵w=z+ai=1+(a-1)i,
∴w/z=(1+"(" a"-" 1")" i)/(1"-" i)=("[" 1+"(" a"-" 1")" i"](" 1+i")" )/2=(2"-" a+ai)/2,
∴|w/z|=√("(" 2"-" a")" ^2+a^2 )/2≤√2,
∴a2-2a-2≤0,
∴1-√3≤a≤1+√3.
故a的取值范围是[1-√3,1+√3].
10.导学号88184060已知复数z1=2+i,2z2=(z_1+i)/("(" 2i+1")-" z_1 ).
(1)求z2;
(2)若△ABC的三个内角A,B,C依次成等差数列,且μ=cos A+2icos2C/2,求|μ+z2|的取值范围.
解(1)z2=(1/2 "[(" 2+i")" +i"]" )/("(" 2i+1")-(" 2+i")" )=(1+i)/(i"-" 1)=2i/("-" 2)=-i.
(2)在△ABC中,∵A,B,C依次成等差数列,
∴2B=A+C,A+B+C=180°.
∴B=60°,A+C=120°.
∵μ+z2=cos A+2icos2C/2-i
=cos A+(2cos^2 C/2 "-" 1)i
=cos A+icos C,
∴|μ+z2|2=cos2A+cos2C
=(1+cos2A)/2+(1+cos2C)/2
=1+1/2(cos 2A+cos 2C)
=1-1/2cos(A-C).
∵A+C=120°,
∴A-C=120°-2C,且0° ∴-120° ∴-1/2 ∴1/2≤1-1/2cos(A-C)<5/4. ∴|μ+z2|的取值范围是[√2/2 "," √5/2).