A) circle
B) straight line
C) parabola
D) none of these
Correct Answer: B
Solution :
Given that \[\frac{2z+1}{iz+1}=-2\] Put, \[z=x+iy\] \[\therefore \] \[\frac{2z+1}{iz+1}=\frac{2(x+iy)+1}{i(x+iy)+1}\] \[=\frac{(2x+1)+2iy}{ix-(y-1)}\times \frac{(ix+(y-1))}{(ix+(y-1))}\] \[=\frac{[(2x+1)x+2y(y-1)]i+(2x+1)(y-1)-2xy}{-{{x}^{2}}-{{(y+1)}^{2}}}\] \[\therefore \] Imaginary part of \[\frac{2z+1}{iz+1}\] \[=\frac{(2x+1)x+2{{y}^{2}}-2y}{-{{x}^{2}}-{{(y-1)}^{2}}}=-2\] \[\Rightarrow \]\[2{{x}^{2}}+x+2{{y}^{2}}-2y=2{{x}^{2}}+2{{y}^{2}}-4y+2\] \[\Rightarrow \]\[x+2y=2\] This represents a equation of straight line.
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