若z为复数,且(2-3i)z=(-1+i)z+2-i,则│z│=_______

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若z为复数,且(2-3i)z=(-1+i)z+2-i,则│z│=_______
若z为复数,且(2-3i)z=(-1+i)z+2-i,则│z│=_______

若z为复数,且(2-3i)z=(-1+i)z+2-i,则│z│=_______
(2-3i)z=(-1+i)z+2-i =>
(2-3i)z-(-1+i)z=2-i,=>
(3-4i)z=2-i =>
(3+4i)(3-4i)z=(2-i)(3+4i) =>
25z=(2-i)(3+4i) =>
z=1/25*(2-i)(3+4i)
|z|^2=1/25^2*(2-i)(3+4i)*(2+i)(3-4i)=1/5
|z|=sqrt(5)/5