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

The Correct Option is A

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

Answer 2

We are given the conditions for the real and imaginary parts of \( z \). Let \( z = x + iy \), where \( x \) and \( y \) are real numbers. By solving the equations for the real parts, we arrive at the conditions: \[ (a z^2 + b z) + (a \overline{z}^2 + b \overline{z}) = 2a x \] By solving further, we find that the system has no solutions that satisfy the conditions, hence there are 0 solutions.