Show that the complex numbers z satisfying `z^(2)+(barz)^(2)=2` constitute a hyperbola.

Explanation

Substituting `z=x+iy` in the given equation `z^(2)+(barz)^(2)=2`, we obtian the cartesiaon form of the given equation.
`:.(x+iy)^(2)+(x-iy)^(2)=2`
i.e., `x^(2)-y^(2)+2i xy+x^(2)-y^(2)-2i xy =2`
i.e., `x^(2)-y^(2)=1`
Since, the equation denotes a hyperbola, all the complex numbers satisfying `z^(2)+( bar z)^(2)=2` lie on the hyperboa `x^(2)-y^(2)=1`.