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If $|z+\bar{z}|=|z-\bar{z}|$, then the locus of $z$ is:
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Verified Answer
The correct answer is:
A pair of straight lines
Let $z=x+$ iy, $\bar{z}=x-i y$
$|\mathrm{z}+\overline{\mathrm{z}}|=|\mathrm{z}-\overline{\mathrm{z}}|$
$|(\mathrm{x}+\mathrm{iy})+(\mathrm{x}-\mathrm{iy})|=|(\mathrm{x}+\mathrm{iy})-(\mathrm{x}-\mathrm{iy})|$
$|2 \mathrm{x}|=|2 \mathrm{y}|$
$\mathrm{x}=\pm \mathrm{y}$
$|\mathrm{z}+\overline{\mathrm{z}}|=|\mathrm{z}-\overline{\mathrm{z}}|$
$|(\mathrm{x}+\mathrm{iy})+(\mathrm{x}-\mathrm{iy})|=|(\mathrm{x}+\mathrm{iy})-(\mathrm{x}-\mathrm{iy})|$
$|2 \mathrm{x}|=|2 \mathrm{y}|$
$\mathrm{x}=\pm \mathrm{y}$
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