Maths · Modulus and argument (or amplitude) of a complex number
In the Argand's plane, the locus of 1) such that \right\}=\frac{2 \pi}{3} i s \)
In the Argand's plane, the locus of \( z(\neq \) 1) such that \( \arg \left\{\frac{3}{2}\left(\frac{2 z^{2}-5 z+3}{3 z^{2}-z-2}\right)\right\}=\frac{2 \pi}{3} i s \)
- A. a hyperbola with the directrices at \( z=-3 / 2 \) and \( z= \) \( -2 / 3 \)
- B. an ellipse with the directrices at \( z=3 / 2 \) and \( z=2 / 3 \)
- C. a segment of a circle subtending angle \( \frac{2 \pi}{3} \) on arc between points \( z=-3 / 2 \) and \( z=2 / 3 \) lying below real axis.
- D. a segment of a circle subtending angle \( \frac{2 \pi}{3} \) on arc between points \( z=3 / 2 \) and \( z=-2 / 3 \) lying above real axis.
Step-by-step solution
Simplify the given expression: factor numerator and denominator, cancel (z-1), get arg( (2z-3)/(3z+2) ) = 2π/3. Since (2z-3)=2(z-3/2) and (3z+2)=3(z+2/3), this becomes arg(z-3/2) - arg(z+2/3) = 2π/3. Interpreting as the angle subtended by points A(3/2) and B(-2/3) at point z, the locus is an arc of a circle with chord AB. The center lies on the perpendicular bisector and is below the real axis (center (5/12, -13/(12√3))). A test point (topmost point at (5/12, 13/(12√3))) gives the required angle and is above the real axis, confirming the arc lies above the real axis. Hence the locus is a segment of a circle between z=3/2 and z=-2/3 lying above the real axis.