Describe the locus of points z satisfying the given equation:
\begin{equation*}
|z+1|^{2}+2|z|^{2}=|z-1|^{2}
\end{equation*}
Answer:
Let \begin{equation*}
z=x+iy
\end{equation*}
Then, we have
\begin{equation*}
|(x+1)+iy|^{2}+2|x+iy|^{2}=|(x-1)+iy|^{2}
\end{equation*}
Thus,
\begin{equation*}
(x+1)^{2}+y^{2}+2(x^{2}+y^{2})=(x-1)^{2}+y^{2}
\end{equation*}
Simplifying the equation and we get,
\begin{equation*}
x^{2}+2x+y^{2}=0
\end{equation*}
Adding 1 at both sides, we obtain
\begin{equation*}
x^{2}+2x+1+y^{2}=1
\end{equation*}
\begin{equation*}
(x+1)^{2}+y^{2}=1
\end{equation*}
Therefore, the locus of point z is a circle centered at (-1,0) with radius 1.
