Question

If $a ≠ ±b$ and are purely real, z ϵ complex number, $Re(az^{2}+bz)=a$ and $Re(bz^{2}+az)=b$ then number of value of z possible is

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (A)

0

Answer 2

$a(x^2-y^2)+bx=a…..(i)$
$b(x^2-y^2)+ax=b…..(ii)$
$(i)-(ii)$
$(a-b)(x^2-y^2)+(b-a)x=a-b\ \ \ \ (a≠b)$
$⇒ x^2-y^2-x=1$
$(i)+(ii)$
$(a+b)(x^2-y^2)+x(a+b)=a+b$ $(a≠-b)$
$⇒ x^2-y^2+x=1$
$⇒x=0$
$⇒y^2=-1$
therefore, no complex number is possible.
The correct option is (A): 0