A) A circle
B) A straight line
C) A parabola
D) None of these
Correct Answer: B
Solution :
We have \[\frac{2z+1}{iz+1}=\frac{2(x+iy)+1}{i(x+iy)+1}=\frac{(2x+1)+2iy}{(1-y)+ix}\] \[=\frac{[(2x+1)(1-y)+2xy]+i[2y(1-y)-x(2x+1)]}{{{(1-y)}^{2}}+{{x}^{2}}}\] But it is given that imaginary part of \[\frac{(2z+1)}{(iz+1)}\]is - 2 Þ\[x+2y-2=0\]. Which is a straight line.You need to login to perform this action.
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