Let a- b- x and y be real numbers such that a-b-1 and y-0- If the complex numbers z-x- iy satisfies Ima z-b-z-1-y- then which of the following is-are possible values of x -A- 1-1- y 2B- -1-1-y2- 1-1- y 2D- -1-1- y 2

The correct options are B 1 √(1 y^2) D 1+√(1 y^2) az+b/z+1=ax+b+aiy/(x+1)+iy =(ax+b+aiy)((x+1) iy)/(x+1)^2 +y^2 ∴ Im(ax+b/z+1)= (ax+b)y+ay(x+1)/(x+1)^2 +y^2 ...

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Byju's Answer

Standard XI

Mathematics

Purely Imaginary

Let a, b, x a...

Question

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

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

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

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

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

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Solution

The correct options are
B
$- 1 - \sqrt{1 - y^{2}}$

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

$\frac{a z + b}{z + 1} = \frac{a x + b + a i y}{(x + 1) + i y} = \frac{(a x + b + a i y) ((x + 1) - i y)}{(x + 1)^{2} + y^{2}} ∴ I m (\frac{a x + b}{z + 1}) = \frac{- (a x + b) y + a y (x + 1)}{(x + 1)^{2} + y^{2}} \Rightarrow \frac{(a - b) y}{(x + 1)^{2} + y^{2}} = y ∵ a - b = 1 ∴ (x + 1)^{2} + y^{2} = 1 ∴ x = - 1 \pm \sqrt{1 - y^{2}}$

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

The correct options are B −1−√1−y2 D −1+√1−y2 az+bz+1=ax+b+aiy(x+1)+iy=(ax+b+aiy)((x+1)−iy)(x+1)2+y2∴Im(ax+bz+1)=−(ax+b)y+ay(x+1)(x+1)2+y2⇒(a−b)y(x+1)2+y2=y∵a−b=1∴(x+1)2+y2=1∴x=−1±√1−y2

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