JEE Advanced 2017 · Paper 1 · Mathematics Q6

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MathematicsMultiple correct+4 / -2Question 42 of 54

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

-1 + \sqrt{1 - y^2}

-1 - \sqrt{1 - y^2}

1 + \sqrt{1 + y^2}

1 - \sqrt{1 + y^2}

Answer and worked solution

Official answer: A, B

Rewrite. Since $a-b=1$ , $az+b=a(z+1)-1$ , so
$\frac{az+b}{z+1}=a-\frac1{z+1}.$

Imaginary part.
$\operatorname{Im}\left(-\frac1{z+1}\right)=\frac y{(x+1)^2+y^2}.$

Solve. Setting this equal to $y$ (with $y\ne0$ ):
$(x+1)^2+y^2=1\implies x=-1\pm\sqrt{1-y^2}.$

(A), (B)

Topics:Algebra of complex numbers Geometry in the complex plane

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JEE Advanced 2017 · Paper 1, Mathematics question 6