已知復數(shù)Z滿足:|Z|=1+3i-Z,求[(1+i)^2(3+4i)^2]/2Z
已知復數(shù)Z滿足:|Z|=1+3i-Z,求[(1+i)^2(3+4i)^2]/2Z
數(shù)學人氣:810 ℃時間:2020-01-30 02:58:17
優(yōu)質(zhì)解答
|Z|=1+3i-Z設(shè)z=x+yi|z|=√(x^2+y^2)|Z|=1+3i-Z,√(x^2+y^2)=(1-x)+(3-y)i∴√(x^2+y^2)=1-x,且3-y=0∴y=3√(x^2+9)=1-x x^2+9=(1-x)^29=1-2x,x=-4,符合題意∴z=-4+3i∴[(1+i)^2(3+4i)^2]/2Z=[2i(-7+24i)]/[2(-4+3i)]...
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