\[z\text{ }=\text{ }x\text{ }+\text{ }iy\] ⇒ z̅ = x – iy

Now, we also have,

z z̅ + 2 (z + z̅) + b = 0

\[\Rightarrow ~\left( x\text{ }+\text{ }iy \right)\text{ }\left( x\text{ }\text{ }iy \right)\text{ }+\text{ }2\text{ }\left( x\text{ }+\text{ }iy\text{ }+\text{ }x\text{ }\text{ }iy \right)\text{ }+\text{ }b\text{ }=\text{ }0\]

\[\Rightarrow ~{{x}^{2}}~+\text{ }{{y}^{2}}~+\text{ }4x\text{ }+\text{ }b\text{ }=\text{ }0\]

The equation obtained represents the equation of a circle.