app输入错误次数过多:i 是虚数单位,i/(1+i)=?
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i
解:i/(1+i)
=i(1-i)/[(1+i)(1-i)]
=(-i^2+i)/(1-i^2)
=(1+i)/(1+1)
=(1/2)+(1/2)i
i^2=-1
i/(1+i)=i(1-i)/(1-i^2)=(1-i)/2
1/1-i
解:i/(1+i)
=i(1-i)/[(1+i)(1-i)]
=(-i^2+i)/(1-i^2)
=(1+i)/(1+1)
=(1/2)+(1/2)i
i^2=-1
i/(1+i)=i(1-i)/(1-i^2)=(1-i)/2
1/1-i