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If a complex number $z$ satisfies
$|z|^2+1=\mid z^2-1$, then the locus of $z$ is
Options:
$|z|^2+1=\mid z^2-1$, then the locus of $z$ is
Solution:
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Verified Answer
The correct answer is:
the imaginary axis
$\begin{aligned} & \text { Let } z=x+i y \text {, so } \\ & \qquad|z|^2+1=\left|z^2-1\right|\end{aligned}$
$$
\begin{aligned}
& \Rightarrow x^2+y^2+1=\sqrt{\left(x^2-y^2-1\right)^2+4 x^2 y^2} \\
& \Rightarrow\left(x^2+y^2+1\right)^2=\left(x^2-y^2-1\right)^2+4 x^2 y^2 \\
& \Rightarrow\left(x^2+y^2+1\right)^2-\left(x^2-y^2-1\right)^2=4 x^2 y^2 \\
& \Rightarrow \quad\left[\left(x^2+y^2+1\right)+\left(x^2-y^2-1\right)\right] \\
& \quad\left[\left(x^2+y^2+1\right)-\left(x^2-y^2-1\right)\right]=4 x^2 y^2 \\
& \Rightarrow\left(2 x^2\right)\left(2 y^2+2\right)=4 x^2 y^2 \\
& \Rightarrow \quad x^2 y^2+2 x^2=x^2 y^2 \Rightarrow x^2=0 \Rightarrow x=0,
\end{aligned}
$$
so locus is a imaginary axis.
$$
\begin{aligned}
& \Rightarrow x^2+y^2+1=\sqrt{\left(x^2-y^2-1\right)^2+4 x^2 y^2} \\
& \Rightarrow\left(x^2+y^2+1\right)^2=\left(x^2-y^2-1\right)^2+4 x^2 y^2 \\
& \Rightarrow\left(x^2+y^2+1\right)^2-\left(x^2-y^2-1\right)^2=4 x^2 y^2 \\
& \Rightarrow \quad\left[\left(x^2+y^2+1\right)+\left(x^2-y^2-1\right)\right] \\
& \quad\left[\left(x^2+y^2+1\right)-\left(x^2-y^2-1\right)\right]=4 x^2 y^2 \\
& \Rightarrow\left(2 x^2\right)\left(2 y^2+2\right)=4 x^2 y^2 \\
& \Rightarrow \quad x^2 y^2+2 x^2=x^2 y^2 \Rightarrow x^2=0 \Rightarrow x=0,
\end{aligned}
$$
so locus is a imaginary axis.
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