Author Topic: Quiz#1 LEC0101 1D  (Read 383 times)

Xun Zheng

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Quiz#1 LEC0101 1D
« on: September 25, 2020, 11:14:04 AM »
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.
« Last Edit: September 25, 2020, 11:20:27 AM by Xun Zheng »