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Q. The locus of point $ z $ satisfying $ \operatorname{Re}\left( \frac{1}{z} \right)=k, $ where $ k $ is a non- zero real number, is:

KEAMKEAM 2004

Solution:

Let $ z=x+iy, $ then $ \operatorname{Re}\left( \frac{1}{z} \right)=k $
$ \Rightarrow $ $ \operatorname{Re}\left( \frac{1}{x+iy} \right)=k $
$ \Rightarrow $ $ \operatorname{Re}\left( \frac{x}{{{x}^{2}}+{{y}^{2}}}-\frac{iy}{{{x}^{2}}+{{y}^{2}}} \right)=k $
$ \Rightarrow $ $ k=\frac{x}{{{x}^{2}}+{{y}^{2}}} $
$ \Rightarrow $ $ {{x}^{2}}+{{y}^{2}}-\frac{1}{k}x=0 $
which is an equation of circle.