Answer to Question #116898 in Algebra for jay

Question

Determine the complex number z which satisfies the equations |z + 3i| = |z + 5 −2i|
and |z − 4i| = |z + 2i| simultaneously.

Answer 1

Given, $\mid$z + 3i$\mid$ = $\mid$z + 5 - 2i$\mid$ ... eq. (1)

& $\mid$z - 4i$\mid$ = $\mid$z + 2i$\mid$ ... eq. (2)

Let z = x + iy.

Substituting the above value of z in equation (1) :

$\mid$x + iy + 3i$\mid$ = $\mid$x + iy + 5 - 2i$\mid$

$\mid$x + i(y + 3)$\mid$ = $\mid$(x + 5) + i(y - 2)$\mid$

$\sqrt{\smash[b]{ x² + (y + 3)²}}$ = $\sqrt{\smash[b]{ (x + 5)² + (y - 2)²}}$

Squaring the above equation, we get,

x² + (y + 3)² = (x + 5)² + (y - 2)²

x² + y² + 9 + 6y = x² + 25 + 10x + y² + 4 - 4y

i.e., x² + y² + 9 + 6y - x² - 25 - 10x - y² - 4 + 4y = 0

y(6 + 4) - 10x + (9 - 25 - 4) = 0

10y - 10x - 20 = 0

i.e., 10y - 10x = 20 ... eq. (3)

Substituting z = x + iy in equation (2) :

$\mid$x + iy - 4i$\mid$ = $\mid$x + iy + 2i$\mid$

$\mid$x + i(y - 4)$\mid$ = $\mid$x + i(y+2)$\mid$

$\sqrt{\smash[b]{ x² + (y - 4)²}}$ = $\sqrt{\smash[b]{ x² + (y + 2)²}}$

Squaring the above equation, we get,

x² + (y - 4)² = x² + (y + 2)²

x² + y² + 16 - 8y = x² + y² + 4 + 4y

i.e., x² + y² + 16 - 8y - x² - y² - 4 - 4y = 0

16 - 4 + y(-8 - 4) = 0

12 - 12y = 0

i.e., 12y = 12

$\therefore$ y = 1

Substituting the above value of y = 1 in equation (3) :

10y - 10x = 20

10(1) - 10x = 20

10 - 10x = 20

i.e., 10x = 10 - 20 = - 10

$\therefore$ x = -1

Therefore, z = -1 + i