find the complex numbers Z satisfying the equation:- |Z-4| / |Z-8|=1 |Z-12 | /|Z-8i|=5/3 - Maths - Complex Numbers and Quadratic Equations

Question

find the complex numbers Z satisfying the equation:- |Z-4|
/ |Z-8|=1 & |Z-12 | /|Z-8i|=5/3

Lalit Mehra · Accepted Answer

$Given, \frac{| z - 4 |}{| z - 8 |} = 1 and \frac{| z - 12 |}{| z - 8 i |} = \frac{5}{3}$

Putting z = x + iy, we get

$\frac{| z - 4 |}{| z - 8 |} = \frac{| x + i y - 4 |}{| x + i y - 8 |} = 1 ∴ | (x - 4) + i y |^{2} = | (x - 8) + i y |^{2} \Rightarrow (x - 4)^{2} + y^{2} = (x - 8)^{2} + y^{2} \Rightarrow x^{2} - 8 x + 16 = x^{2} - 16 x + 64 \Rightarrow 8 x = 48 \Rightarrow x = 6$

$\frac{| z - 12 |}{| z - 8 i |} = \frac{5}{3} ∴ \frac{| x + i y - 12 |}{| x + i y - 8 i |} = \frac{5}{3} \Rightarrow 3 | (x - 12) + i y | = 5 | x + i (y - 8) | \Rightarrow 9 | (x - 12) + i y |^{2} = 2 5 | x + i (y - 8) |^{2} \Rightarrow 9 [(x - 12)^{2} + y^{2}] = 2 5 [x^{2} + {(y - 8)}^{2}]$

Putting x = 6, we have

$9 [(6 - 12)^{2} + y^{2}] = 25 [(6)^{2} + (y - 8)^{2}] ∴ 9 (36 + y^{2}) = 25 (36 + y^{2} - 6 y + 64) \Rightarrow 324 + 9 y^{2} = 2500 + 25 y^{2} - 400 y \Rightarrow 16 y^{2} - 400 y + 2176 = 0 \Rightarrow y^{2} - 25 y + 136 = 0 \Rightarrow (y - 8) (y - 17) = 0 \Rightarrow y = 8 or y = 17$

When x = 6 and y = 8, we have

z = 6 + 8i

When x = 6 and y = 17, we have

z = 6 + 17i

Thus, z ₁ = 6 + 8i and z ₂ = 6 + 17i are two complex numbers satisfying the given equations.

find the complex numbers Z satisfying the equation:- |Z-4| / |Z-8|=1 & |Z-12 | /|Z-8i|=5/3