Let $z = x + i y$ and $| (2 z - 3) | = | (z - 6) |$ then prove that $x^{2} + y^{2} = 3$ .

Open in App

Solution

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

Suggest Corrections

Similar questions

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

z = x + i y

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

x + y + z = x y z

\frac{2 x}{1 - x^{2}} + \frac{2 y}{1 - y^{2}} + \frac{2 z}{1 - z^{2}} = \frac{2 x}{1 - x^{2}} \cdot \frac{2 y}{1 - y^{2}} \cdot \frac{2 z}{1 - z^{2}}

z

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

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

z = x + i y

| e^{z} | = e^{x}

z = x + i y

\frac{z - 1}{z + 1}

| z | = 1

Join BYJU'S Learning Program