Let a,b,x and y be real numbers such that a−b=1 and y≠0. If the complex number z=x+iy satisfies Im(az+bz+1)=y, then when of the following is (are) possible value(s) of x?
1−√1+y2
−1+√1−y2
1+√1+y2
−1−√1−y2
A
1−√1+y2
B
1+√1+y2
C
−1+√1−y2
D
−1−√1−y2
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Solution
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Im(az+bz+1)=y
Im((ax+b+iay)(x+1−iy)(x+1+iy)(x+1−iy))=y
Im((ax+b+iay)(x+1−iy)(x+1)2+y2)=y
y[(x+1)2+y2]=−y(ax+b)+ay(x+1)=−by+ay=(a−b)y=y
∴y[(x+1)2+y2]=y
∴(x+1)2=1−y2
∴(x+1)=±√1−y2
x=−1±√1−y2
Hence, options B and D are correct.
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