Question

The complex numbers z = x + iy which satisfy the equation $|$ $\frac{z-4i}{z+4i}$ $|$ = 1, lie on

• A. Circle
• B. z + = 0
• C. z - = 0
• D. x - axis

Answer 1

Answer: (C), (D)

The correct options are
C

z - = 0

D

x - axis

$|z-4i|$ = $|z+4i|$
$|x+iy-4i|$ = $|x+iy+4i|$
$\Rightarrow \sqrt{x^2+(y-4{)}^2}$ = $\sqrt{x^2+(y+4{)}^2}$
$\Rightarrow$ $x^2$ + $y^2$ - 8y +16 = $x^2$ + $y^2$ + 8y + 16
16y = 0
y = 0
$\Rightarrow$ x - axis - (d)
option (b):
z + $\overline{z}$ = $x+iy+x-iy$
$\Rightarrow$ x = 0 - not correct
option (c):
z - $\overline{z}$ = 0 $\Rightarrow$ y = 0 - correct

$|$z - $z_1$ $|$= $|$z - $z_2$ $|$
$\Rightarrow$ z lies on the perpendicular bisector of $z_1$ and $z_1$
In this case $z_1$ = - 5i, $z_2$ = +5i
$\Rightarrow$ The perpendicular bisectors is x - axis