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复数z={(1+i)²+3(1-i)}/(2+i)若z²+(a)/(z)
题目详情
复数z={(1+i)²+3(1-i)}/(2+i)若z²+(a)/(z)<0求纯虚数a
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答案和解析
z
=[(1+i)^2+3(1-i)]/(2+i)
=(1+2i+i^2+3-3i)/(2+i)
=(1+2i-1+3-3i)/(2+i)
=(3-i)/(2+i)
=(3-i)(2-i)/(4-i^2)
=(6-5i+i^2)/(4+1)
=(6-5i-1)/5
=1-i,
∴z^2=1-2i+i^2=1-2i-1=-2i,
∴z^2+a/z=-2i+a/(1-i)=-2i+a(1+i)/(1-i^2)=-2i+a/2+(a/2)i<0.
∵a是纯虚数,∴-2i+a/2=0,∴-4i+a=0,∴a=4i.
=[(1+i)^2+3(1-i)]/(2+i)
=(1+2i+i^2+3-3i)/(2+i)
=(1+2i-1+3-3i)/(2+i)
=(3-i)/(2+i)
=(3-i)(2-i)/(4-i^2)
=(6-5i+i^2)/(4+1)
=(6-5i-1)/5
=1-i,
∴z^2=1-2i+i^2=1-2i-1=-2i,
∴z^2+a/z=-2i+a/(1-i)=-2i+a(1+i)/(1-i^2)=-2i+a/2+(a/2)i<0.
∵a是纯虚数,∴-2i+a/2=0,∴-4i+a=0,∴a=4i.
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