Let a,b,x and y be real numbers such that a−b=1 and y≠0. If the complex numbers z=x+iysatisfies Im (az+bz+1)=y, then which of the following is possible value of x?
A
1−√1+y2
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B
−1−√1−y2
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C
1+√1+y2
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D
−1+√1−y2
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Solution
The correct option is D−1+√1−y2 az+bz+1=ax+b+aiy(x+1)+iy=(ax+b+aiy)((x+1)−iy)(x+1)2+y2∴Im(ax+bz+1)=−(ax+b)y+ay(x+1)(x+1)2+y2⇒(a−b)y(x+1)2+y2=y∵a−b=1∴(x+1)2+y2=1∴x=−1±√1−y2