Let z = x + iy
Now,
![\[\Rightarrow ~{{x}^{2}}~+\text{ }x\text{ }\text{ }-2\text{ }+\text{ }{{y}^{2}}~=\text{ }4\text{ }\left( {{x}^{2}}~\text{ }-2x\text{ }+\text{ }1\text{ }+\text{ }{{y}^{2}} \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-7ad37dcee3d2672346303953813c2676_l3.png "Rendered by QuickLaTeX.com")
![\[\Rightarrow ~3{{x}^{2}}~+\text{ }3{{y}^{2}}~\text{ }-9x\text{ }+\text{ }6\text{ }=\text{ }0\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-aa6e20d5093775bc98a53049feee223b_l3.png "Rendered by QuickLaTeX.com")
The equation obtained represents the equation of a circle.
Hence, locus of z is a circle.
If the real part of ( z̅ + 2)/ ( z̅ – 1) is 4, then show that the locus of the point representing z in the complex plane is a circle.
If the real part of ( z̅ + 2)/ ( z̅ – 1) is 4, then show that the locus of the point representing z in the complex plane is a circle.