Prove that $x^{2} + y^{2} = 9$ where $z = x + i y$ and $| z + 6 | = | 2 z + 3 |$ .

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Solution

Given $z = x + i y$ .
Also given
$| z + 6 | = | 2 z + 3 |$
or, $| (x + 6) + i y | = | (2 x + 3) + 2 i y |$
or, $\sqrt{(x + 6)^{2} + (y)^{2}} = \sqrt{(2 x + 3)^{2} + (2 y)^{2}}$
or, $(x + 6)^{2} + y^{2} = (2 x + 3)^{2} + 4 y^{2}$
or, $3 x^{2} + 3 y^{2} = 27$
or, $x^{2} + y^{2} = 9$ .

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