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Q.

Let  a,b,x  and y be real numbers such that  a−b=1and y≠0. If the complex number  z=x+iy satisfies Imaz+bz+1=y, then which of the following is (are) possible value of x?

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a

−1+1−y2

b

1−1+y2

c

1+1+y2

d

−1−1−y2

answer is A.

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Detailed Solution

az+bz+1−az¯+bz¯+1=z−z¯⇒azz¯+az+bz¯+b−azz¯−bz−az¯−b=z−z¯z+1z¯+1⇒az +bz¯−bz−az¯=z−z¯z+1z¯+1⇒a−bz−z¯=z−z¯z+1z¯+1since y≠0⇒ z−z¯≠0 ⇒ z+1z¯+1=1x+12+y2=1⇒x+1=±1−y2⇒x=−1±1−y2
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