已知复数Z满足Z+1/Z∈R,且(Z-2)的模=2,求Z

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已知复数Z满足Z+1/Z∈R,且(Z-2)的模=2,求Z

已知复数Z满足Z+1/Z∈R,且(Z-2)的模=2,求Z
已知复数Z满足Z+1/Z∈R,且(Z-2)的模=2,求Z

已知复数Z满足Z+1/Z∈R,且(Z-2)的模=2,求Z
设z=x+yi
z+1/z=(x+yi)+1/(x+yi)=(x+yi)+(x-yi)/(x²+y²)=x+x/(x²+y²)+[y-y/(x²+y²)]i∈R,y-y/(x²+y²)=0,得x²+y²=1
|z-2|=|x+yi-2|=|(x-2)+yi|=√[(x-2)²+y²]=2,(x-2)²+y²=4
解此方程组,得x=1/4,y=±√15/4
综上,z=(1/4)+(√15/4)i或z=(1/4)-(√15/4)i