Question

Let a, b, x and y be real numbers such that a - b = 1 and y ≠ 0. If the complex number z = x + iy satisfies
Im (az+b/z+1) = y, then which of the following is(are) possible value(s) of x ?
[A] -1-√1-y²
[B] 1+√1+y²
[C] 1-√1+y²
[D] -1+√1-y²

Answer 1

ANSWER:

• Options A) and D) are correct.

GIVEN:

• a, b, x and y be real numbers.

• a - b = 1 and y ≠ 0.

• The complex number z = x + iy satisfies Im {(az+b) / (z+1)} = y.

TO FIND:

• The possible values of x.

EXPLANATION:

$\sf \dashrightarrow Im\bigg[ \dfrac{az+b}{ z+1}\bigg] = y$

$\bf We\ know\ that \ z = x + iy$

$\sf \dashrightarrow Im\bigg[\dfrac{a( x + iy)+b}{ x + iy+1}\bigg] = y$

$\sf \dashrightarrow Im\bigg[ \dfrac{ax + aiy+b}{ x + 1 + iy} \bigg]= y$

$\sf \dashrightarrow Im\bigg[ \dfrac{ax + aiy+b}{ x + 1 + iy} \times \frac{x+ 1 - iy}{x+ 1 - iy} \bigg]= y$

$\sf \dashrightarrow Im\bigg[ \dfrac{(ax + aiy+b)(x+ 1 - iy)}{( x + 1 {)}^{2} - (iy {)}^{2} }\bigg]= y$

$\sf \dashrightarrow \dfrac{ - y(ax + b) + ay(x + 1)}{( x + 1 {)}^{2} - (- y^{2}) }= y$

$\sf \dashrightarrow \dfrac{ -(ax + b) + a(x + 1)}{( x + 1 {)}^{2} - (- y^{2}) }= 1$

$\sf \dashrightarrow \dfrac{ -ax - b + ax + a}{( x + 1 {)}^{2} - (- y^{2}) }= 1$

$\sf \dashrightarrow \dfrac{ a-b}{( x + 1 {)}^{2} - (- y^{2}) }= 1$

$\bf We\ know\ that \ a - b = 1$

$\sf \dashrightarrow \dfrac{ 1}{( x + 1 {)}^{2} - (- y^{2}) }= 1$

$\sf \dashrightarrow (x + 1 {)}^{2} + y^{2}= 1$

$\sf \dashrightarrow (x + 1 {)}^{2} = 1 - y^{2}$

$\sf \dashrightarrow x + 1 = \pm \sqrt{ 1 - y^{2}}$

$\sf \dashrightarrow x = - 1 \pm \sqrt{ 1 - y^{2}}$

$\bf Hence\ the\ value \ of \ x = - 1 \pm \sqrt{ 1 - y^{2}}$