Let $a, b, x$ and $y$ be real numbers such that $a-b=1$ and $y \neq 0$. If the complex number $z=x+i y$ satisfies $\operatorname{Im}\left(\frac{a z+b}{z+1}\right)=\mathrm{y}$, then which of the following is(are) possible value(s) of $x$?

(A) $-1+\sqrt{1-y^{2}}$; (B) $-1-\sqrt{1-y^{2}}$

JEE Advanced · 2017 · 21 May 2017 · Paper 1 · Complex Numbers

Let

a, b, x

and

y

be real numbers such that

a - b = 1

and

y \neq = 0

. If the complex number

z = x + i y

satisfies

Im (\frac{a z + b}{z + 1}) = y

, then which of the following is(are) possible value(s) of

x

Answer: (A), (B)

We have

Im (\frac{a ( x + i y ) + b}{( x + i y ) + 1}) = y

. Multiplying numerator and denominator by the conjugate of the denominator and simplifying, the imaginary part condition gives

(x + 1)^{2} + y^{2} = 1

. Solving for

x

x = - 1 \pm 1 - y^{2}

JEE Advanced · 2017 · 21 May 2017 · Paper 1 · One or more correct options