For $z$ being a complex show that $| z + 6 | = | 2 z + 3 |$ gives $x^{2} + y^{2} = 9$ .

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Solution

Given,
$| z + 6 | = | 2 z + 3 |$
or, $| (x + 6) + i y | = | (2 x + 3) + I (2 y) |$
or, $\sqrt{(x + 6)^{2} + y^{2}} = \sqrt{(2 x + 3)^{2} + (2 y)^{2}}$
or, $(x + 6)^{2} + y^{2} = (2 x + 3)^{2} + (2 y)^{2}$
or, $x^{2} + 12 x + 36 + y^{2} = 4 x^{2} + 12 x + 9 + 4 y^{2}$
or, $3 x^{2} + 3 y^{2} = 27$
or, $x^{2} + y^{2} = 9$ .

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Similar questions

x^{2} + y^{2} = 9

z = x + i y

| z + 6 | = | 2 z + 3 |

z = x + i y

| (2 z - 3) | = | (z - 6) |

x^{2} + y^{2} = 3

z = a + i b

z = {(a + i b)}^{2}

x^{2} + y^{2} = {(a^{2} + b^{2})}^{2}

z = a x^{2} + 2 h x y + b y^{2}

x^{2} \frac{\partial^{2} z}{\partial x^{2}} + 2 x y \frac{\partial^{2} z}{\partial x \partial y} + y^{2} \frac{\partial^{2} z}{\partial y^{2}} = ?

z = x + i y

| z - 1 | + | z + 1 | \leq 4

z

\frac{x^{2}}{4} + \frac{y^{2}}{3} = 1

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