A) ellipse
B) parabola
C) straight line
D) circle
Correct Answer: D
Solution :
Let \[z=x+iy\] \[\therefore \] \[\frac{z-1}{2z+1}=\frac{x+iy-1}{2x+2iy+1}\] \[=\frac{(x-1)+iy}{(2x+1)+2iy}\times \frac{(2x+1)-2iy}{(2x+1)-2iy}\] \[=\frac{(x-1)(2x+1)-2iy(x-1)+iy(2x+1)+2{{y}^{2}}}{{{(2x+1)}^{2}}+4{{y}^{2}}}\] \[=\frac{\{(x-1)(2x+1)+2{{y}^{2}}\}+iy\{-2x+2+2x+1\}}{{{(2x+1)}^{2}}+4{{y}^{2}}}\] According to question \[{{I}_{m}}\left( \frac{z-1}{2z+1} \right)=-4\] \[\therefore \] \[\frac{3y}{{{(2x+1)}^{2}}+4{{y}^{2}}}=-4\] \[\Rightarrow \]\[-\frac{3y}{4}=4{{x}^{2}}+4{{y}^{2}}+4x+1\] \[\Rightarrow \]\[16{{x}^{2}}+16{{y}^{2}}+16x+3y+4=0\] This equation represents a circle. \[\therefore \] The locus of z is a circle.
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