2 + 6i = 4z + 8z'

Le z = a+ bi

Then, z' = a-bi

Let us substitutue:

2 + 6i = 4 (a+bi) + 8(a-bi)

Open brackets:

2 + 6i = 4a + 4bi + 8a - 8bi

Group similar terms:

2 + 6i = 4a + 8a + 4bi - 8bi

2 + 6i = 12a - 4bi

Then,

12a= 2 ==> a = 2/12= 1/6

-4b = 6 ==> b = -6/4 = -3/2

Then:

z = (1/6) - (3/2)i

lzl = sqrt(a^2 + b^2)

= sqrt(1/36 + 9/4)

= sqrt(82/36)

= sqrt(82)/6

**lzl = sqrt(82) / 6**

|If z = a+bi, then |z) = |a+bi| = sqrt(a^2+b^2)

Given equation 2+6i = 4z+8z'

Let us assume z = a+bi, then z' = a-bi by definition.

So the given equation becomes:

2+6i = 4(a+bi) +8(a-bi)

2+6i = 12a+(4b-8b)i = 12a -4bi

Equate reals on both sides and imaginaries on both sides:

2 = 12a and

6 = -4b

So a = 1/6 and b = -6/4 = - 3/2

Therefore |Z| = sqrt{( 1/6)^2+(-3/2)^2} = sqrt{1/36+9/4} = sqrt{(1+81)/36} = {sqrt(82)}/6

The expression of the module of a complex number:

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

For the complex number z, written algebraically:

z = a + b*i, we'll have

the real part = Re(z) = a

and

the imaginary part is Im(z) = b.

|z| = sqrt(a^2 + b^2)

The complex number z' is the conjugate of z and it's expression is z' = a - b*i

Now, we'll re-write the given expression:

2+6i=4z+8z'

We'll factorize by 2, both sides:

2(1+3i) = 2(2z+4z')

We'll divide by 2 both sides:

1+3i = 2z+4z'

We'll substitute z and z' into the expression above:

1+3i = 2a+2b*i + 4a - 4b*i

We'll combine like terms from the right side:

1+3i = 6a - 2b*i

The real part from the left side has to be equal to the real part from the right side.

6a = 1

We'll divide by 6:

**a = 1/6**

The imaginary part from the left side has to be equal to the imaginary part from the right side.

3*i= -2b*i

3=-2b

We'll divide by -2:

**b = -3/2**

The module of the complex number z is:

|z| = sqrt (1/36 + 9/4)

|z| = sqrt (82/36)

**|z| = [sqrt (82)] / 6**