设复数z=(a-5)/(a^2+4a-5)+(a^2+2a-15)i为实数,求实数a的值
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![设复数z=(a-5)/(a^2+4a-5)+(a^2+2a-15)i为实数,求实数a的值](/uploads/image/z/3878129-65-9.jpg?t=%E8%AE%BE%E5%A4%8D%E6%95%B0z%3D%28a-5%29%2F%28a%5E2%2B4a-5%29%2B%28a%5E2%2B2a-15%29i%E4%B8%BA%E5%AE%9E%E6%95%B0%2C%E6%B1%82%E5%AE%9E%E6%95%B0a%E7%9A%84%E5%80%BC)
设复数z=(a-5)/(a^2+4a-5)+(a^2+2a-15)i为实数,求实数a的值
设复数z=(a-5)/(a^2+4a-5)+(a^2+2a-15)i为实数,求实数a的值
设复数z=(a-5)/(a^2+4a-5)+(a^2+2a-15)i为实数,求实数a的值
虚部须为0,得:a^2+2a-15=0
(a+5)(a-3)=0
a=-5或-3
同时实部的分母不能为0,即a^2+4a-5≠0,即(a+5)(a-1)≠0,得:a≠-5,1
所以只能取a=-3.
z=(a-5)/(a^2+4a-5)+(a^2+2a-15)i为实数
∴虚部a^2+2a-15=0
∴a=-5或a=3
检验一下
a=-5时分母出现了0,应舍去
∴a=3
是z=(a-5)/(a^2+4a-5)+(a^2+2a-15)i,还是z=(a-5)/[(a^2+4a-5)+(a^2+2a-15)i]?
应该是后者吧?
如果是的话:a=3
z=(a-5)/[(a^2+4a-5)+(a^2+2a-15)i]
z=(a-5)/[(a+5)(a-1)+(a+5)(a-3)i]
z=[(a-5)/(a+5)...
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是z=(a-5)/(a^2+4a-5)+(a^2+2a-15)i,还是z=(a-5)/[(a^2+4a-5)+(a^2+2a-15)i]?
应该是后者吧?
如果是的话:a=3
z=(a-5)/[(a^2+4a-5)+(a^2+2a-15)i]
z=(a-5)/[(a+5)(a-1)+(a+5)(a-3)i]
z=[(a-5)/(a+5)]{1/[(a-1)+(a-3)i]}
z=[(a-5)/(a+5)]{[(a-1)-(a-3)i]/{[(a-1)+(a-3)i][(a-1)-(a-3)i]}}
z=[(a-5)/(a+5)]{[(a-1)-(a-3)i]/[(a-1)^2+(a-3)^2]}
z=[(a-5)/(a+5)]{[(a-1)-(a-3)i]/(2a^2-8a+10)}
z=(a-5)(a-1)/[2(a+5)(a-5)(a+1)]-(a-5)(a-3)i/[2(a+5)(a-5)(a+1)]
z=(a-1)/[2(a+5)(a+1)]-(a-3)i/[2(a+5)(a+1)]
依题意,有:
(a-3)/[2(a+5)(a+1)]=0
解得:a=3
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