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Question

Let a,b,x and y be real numbers such that ab=1 and y0. 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?

A
11+y2
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B
1+1y2
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C
1+1+y2
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D
11y2
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Solution

The correct options are
B 1+1y2
D 11y2
Im(az+bz+1)=y
Im((ax+b+iay)(x+1iy)(x+1+iy)(x+1iy))=y
Im((ax+b+iay)(x+1iy)(x+1)2+y2)=y
y[(x+1)2+y2]=y(ax+b)+ay(x+1) =by+ay =(ab)y=y
y[(x+1)2+y2]=y
(x+1)2=1y2
(x+1)=±1y2
x=1±1y2
Hence, options B and D are correct.

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