A) a straight line
B) a circle
C) an ellipse
D) a hyperbola
Correct Answer: B
Solution :
Let \[z=x+iy,\]then \[\frac{1}{z}=\frac{1}{x+iy}=\frac{x-iy}{(x+iy)(x-iy)}\] \[=\frac{x-iy}{{{x}^{2}}+{{y}^{2}}}=\frac{x}{{{x}^{2}}+{{y}^{2}}}-\frac{iy}{{{x}^{2}}+{{y}^{2}}}\] \[\therefore \]\[\operatorname{Re}\left( \frac{1}{z} \right)=\frac{x}{{{x}^{2}}+{{y}^{2}}}\]But\[\operatorname{Re}\left( \frac{1}{z} \right)=k\] \[\therefore \] \[\frac{x}{{{x}^{2}}+{{y}^{2}}}=k\] \[\Rightarrow \]\[{{x}^{2}}+{{y}^{2}}-\frac{1}{k}x=0\] Which is an equation of a circle. Hence, the required locus is a circle.
You need to login to perform this action.
You will be redirected in
3 sec
