一道复数题目.如果复数Z1的的幅角π/4,且z²共轭-2/z1是实数,（1）求复数Z；（2)|z|=2|z2|,且ARG2=ARG1+π/2求复数Z2;如果复数Z1的幅角是π/4,且z1²共轭-2/z1是实数，（1）求复数Z1；（2)|z1|=2|z2|,且ARG2=ARG1+π/2求复数Z2;
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设z1=r(cosπ/4+isinπ/4),r>0z²1=r²(cosπ/2+isinπ/2)=r²i
z²1共轭=-r²i
2/z1=2/r(cosπ/4-isinπ/4)
=√2r-√2i/r
∴z1²共轭-2/z1
=-r²i-√2r+√2i/r
=-√2r+(√2/r-r²)i∈R
∴√2/r-r²=0
∴r³=√2,r=2^(1/6)
∴z1=2^(1/6)(cosπ/4+isinπ/4)
(2)∵|z1|=2|z2|,
∴|z2|=1/2|z1|=1/2*2^(1/6)=2^(-5/6)
∵ARGz2=ARGz1+π/2=3π/4
∴z2=2^(-5/6)(cos3π/4+isin3π/4)