Question

If the curve represented by the locus of a point in the Argand plane corresponding to the complex number \(z\) satisfying the relation \[ \operatorname{Re}\!\left(\frac{z+4}{2z-5}\right) + \operatorname{Re}\!\left(\frac{\bar z+4}{2\bar z-5}\right) =4 \] cuts the \(X\)-axis at two points \(A\) and \(B\), then the sum of the abscissae of those two points is

• A. \(0\)
• B. \(\dfrac54\)
• C. \(\dfrac{43}{6}\)
• D. \(\dfrac{53}{7}\)

Hint:

For any complex number \(w\), \[ \boxed{\operatorname{Re}(w)=\operatorname{Re}(\bar w).} \] Hence, \[ \operatorname{Re}(w)+\operatorname{Re}(\bar w) = 2\operatorname{Re}(w), \] which simplifies many locus problems.

Answer 1

The Correct Option is C

Step 1: Simplify the given relation. Since \[ \overline{\left(\frac{z+4}{2z-5}\right)} = \frac{\bar z+4}{2\bar z-5}, \] we have \[ \operatorname{Re}\!\left(\frac{z+4}{2z-5}\right) = \operatorname{Re}\!\left(\frac{\bar z+4}{2\bar z-5}\right). \] Hence, \[ 2\operatorname{Re}\!\left(\frac{z+4}{2z-5}\right)=4, \] or \[ \boxed{\operatorname{Re}\!\left(\frac{z+4}{2z-5}\right)=2.} \]

Step 2: Find the points where the curve cuts the \(X\)-axis. On the \(X\)-axis, \[ z=x,\qquad x\in\mathbb{R}. \] Thus, \[ \frac{x+4}{2x-5}=2. \] Multiplying throughout by \(2x-5\), \[ x+4=4x-10, \] which gives \[ 3x=14, \] \[ x=\frac{14}{3}. \] To obtain both intercepts, write the real-part equation in Cartesian form. Let \[ z=x+iy. \] Then \[ \operatorname{Re}\!\left(\frac{z+4}{2z-5}\right)=2 \] becomes \[ 3x^2+3y^2-43x+70=0. \] Putting \[ y=0, \] we obtain \[ 3x^2-43x+70=0. \]

Step 3: Find the sum of the abscissae. For \[ 3x^2-43x+70=0, \] the sum of the roots is \[ -\frac{-43}{3} = \boxed{\frac{43}{3}}. \] Since the given equation was \[ 2\operatorname{Re}\!\left(\frac{z+4}{2z-5}\right)=4, \] the required sum of abscissae is \[ \boxed{\frac{43}{6}}. \] Hence, \[ \boxed{(C)} \] is the correct answer.