(2-3i)z=(-1+i)z+2-i 求|z|
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(2-3i)z=(-1+i)z+2-i求|z|(2-3i)z=(-1+i)z+2-i求|z|(2-3i)z=(-1+i)z+2-i求|z|令z=a+bi带入得:2a+3b+(2b-3a)i=-a-b+
(2-3i)z=(-1+i)z+2-i 求|z|
(2-3i)z=(-1+i)z+2-i 求|z|
(2-3i)z=(-1+i)z+2-i 求|z|
令z=a+bi
带入得:2a+3b+(2b-3a)i=-a-b+(a-b)i+2-i
即2a+3b=-a-b+2、2b-3a=a-b-1
得a=2/5,b=1/5
所以|z|=1/根号5
解方程得x=-2/5-1/5*i
|z|=sqrt[(2/5)^2+(1/5)^2]=sqrt(5)/5
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