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数学
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已知复数z满足:lzl=1+3i-z,求(1+i)^2(3+4i)^2/2z的值
如题
人气:343 ℃ 时间:2020-03-29 17:53:26
解答
设z=x+iy,x,y为实数.由题意 lzl=1+3i-z,等式左边模长为实数,故等式右边虚数部分必抵消,否则等式不成立,可判断,y=-3,即z=x-3i,代回原等式,x^2+9=(x+1)^2,解得x=4
故z=4-3i
(1+i)^2(3+4i)^2/2z
=2i*(3+4i)^2/[2*(4-3i)]
=i*(3+4i)^2/(4-3i)
=i*(3+4i)(3+4i)(4+3i)/25
=(3+4i)(3+4i)(4i-3)/25
=(3+4i)*(-25)/25=-3-4i
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