if |z-4|/|z-8|=1 then what is [Re(z)] ¿
Re(z) is the real part of the complex number z.
|z-4|/|z-8|=1
|z-4|=|z-8|
let z = x + yi
|x+ yi - 4| = |x+yi - 8|
|(x-4) + yi| = |(x-8) + yi|
according to the above equation, the modulus of the 2 complex numbers are equal
√{(x-4)^2 + y^2} = √{(x-8)^2 + y^2}
x^2 - 8x + 16 +y^2 = x^2 - 16x + 64 + y^2
-8x +16 = -16x + 64
8x = 48
x = 6
therefore
z = 6 + yi
the real part is 6
so
Re(z) = 6
z = x + yi => |z| = â(x^2 + y^2) = â((Re(z))^2 + (Im(z))^2)
|z - 4|/|z - 8| = 1
(x - 4)^2 + y^2= (x - 8)^2 + y^2
x^2 - 8x + 16 = x^2 - 16x + 64
8x = 48 => x = 6 = Re(z).
Comments
Re(z) is the real part of the complex number z.
|z-4|/|z-8|=1
|z-4|=|z-8|
let z = x + yi
|x+ yi - 4| = |x+yi - 8|
|(x-4) + yi| = |(x-8) + yi|
according to the above equation, the modulus of the 2 complex numbers are equal
√{(x-4)^2 + y^2} = √{(x-8)^2 + y^2}
x^2 - 8x + 16 +y^2 = x^2 - 16x + 64 + y^2
-8x +16 = -16x + 64
8x = 48
x = 6
therefore
z = 6 + yi
the real part is 6
so
Re(z) = 6
z = x + yi => |z| = â(x^2 + y^2) = â((Re(z))^2 + (Im(z))^2)
|z - 4|/|z - 8| = 1
(x - 4)^2 + y^2= (x - 8)^2 + y^2
x^2 - 8x + 16 = x^2 - 16x + 64
8x = 48 => x = 6 = Re(z).