# Determine the absolute value of complex number z if z-2z'=2-4i? z=x+i*y z'=x-i*y

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Given that:

z= x+ iy

z' = x - iy

We need to find the absolute value of z such that:

z - 2z' = 2- 4i

Let us substitute with z and z'.

==> (x+ yi) -2 ( x-yi) = 2-4i

==> x + yi - 2x + 2yi = 2- 4i

Now we will combine like terms.

==> (x-2x) + (yi + 2yi) = 2- 4i

==> -x + 3y*i = 2- 4i

==> -x = 2 ==> x = -2

==> 3y = -4 ==> y= -4/3

==> z = -2 - (4/3)*i

Now we will calculate the absolute value l zl .

We know that:

l z l = sqrt(x^2 + y^2)

==> l zl = sqrt( -2^2 + (-4/3)^2

= sqrt( 4 + 16/9) = sqrt( 36+16)/9 = sqrt52/9 =

**Then the absolute value of z is l zl = sqrt(52/9)**

We have z - 2z' = 2 - 4i

As z = x + i*y and z' = x - i*y

z - 2z' = 2 - 4i

=> x + i*y - 2*(x - i*y) = 2 - 4*i

=> x + i*y - 2*x + 2*i*y = 2 - 4*i

=> - x +3*i*y = 2 - 4*i

equate the real and imaginary parts

=> x = -2 and y = -4/3

The required number z is -2 - 4i/3

The absolute value of z is

|z| = sqrt( (-2)^2 + (4/3)^2)

=> sqrt (4 + 16/9)

=> sqrt ( 52/9)

=> (sqrt 52)/3

**The required absolute value is (sqrt 52) / 3**

First, we'll have to determine the complex number z and then, we'll determine the absolute value, |z|.

z - 2z' = 2 - 4i

We'll substitute z and z' by their expressions:

x + i*y - 2(x - i*y) = 2 - 4i

x + i*y - 2x + 2i*y = 2 - 4i

We'll combine the real parts and the imaginary parts:

(x - 2x)+i*(y + 2y) = 2 - 4i (from enunciation)

Comparing, we'll get:

x - 2x=2

-x = 2

x = -2

y + 2y = -4

3y = -4

y = -4/3

The complex number is: z=-2-4i/3

It's absolute value is:

|z| = sqrt[Re(z)^2 + Im(z)^2]

|z| = sqrt[(-2)^2 + (-4/3)^2]

|z| = sqrt ((4 + 16)/9)

|z| = sqrt ((36+ 16)/9)

|z| = sqrt (52/9)

**The absolute value of the complex number is |z| = sqrt (52/9).**