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 which of the following is(are) possible value(s) 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 options are A−1+√1−y2 B−1−√1−y2 a−b=1 and z=x+iy (az+bz+1)=(ax+b)+i(ay)(x+1)+iy={(ax+b)+i(ay)}{(x+1)−iy}(x+1)2+y2 Im(az+bz+1)=ay(x+1)−y(ax+b)(x+1)2+y2⇒y=ay(x+1)−y(ax+b)(x+1)2+y2⇒y=y(a−b)(x+1)2+y2⇒1=1(x+1)2+y2 x=−1±√1−y2