8z + 40z' = 25 + 35i

Let z = a+ bi

==> z' = (a-bi)

8z + 40z' = 25 + 35i

8(a+ bi) + 40(a-bi) = 25 + 35i

8a + 8bi + 40a - 40bi = 25 + 35i

Group similar terms:

48a - 32bi = 25 + 35i

==> 48a = 25

==> a = 25/48

==> -32b = 35

==> b = -35/32

But we know that:

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

= sqrt[(25/48)^2 + (-35/32)^2]

= sqrt13525/ 96

To find |z| if 8z +40z' = 25+35i

Solution:

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

So 8z+40z' = 25+35i

8(a+bi) + 40(a-bi) = 25+35i

8a+8bi+40a-40bi = 25+35i. Collect real umnknowns tohether and imaginary unknowns together on the left:

8a+40a +8bi-40bi = 25 + 35i

48a - 32bi = 25 +35i.

Equate reals and imaginaries respectively on both sides:

48a = 25............(1)

-32bi = 35i..........(2)

So from (1), a = 25/48

From (2), b = -35/32.

So z = a+bi = 25/48 -35/32i

Therefore, |z| = |a+bi| = sqrt (a^2+b^2) = sqrt{(25/48)^2 + (-35/32)^2}

= sqrt {(25*32)^2 +(48*35)^2 / (48*32)^2} = {1/(48*32) }* (1860.752536) = 1.21142

We'll write the expression of the module of a complex number:

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

If z has is written algebraically

z = a + b*i, then

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:

8z+40z'=25+35i

8(a+bi) + 40(a-bi) = 25+35i

We'll remove the brackets:

8a + 8bi + 40a - 40bi = 25+35i

We'll combine like terms from the right side:

48a - 32bi = 25+35i

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

48a=25

We'll divide by 48:

a = 25/48

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

-32b = 35

We'll divide by -32:

b = -35/32

The module of the complex number z = 25/48 - 35*i/32 is:

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

|z| = sqrt[(25/48)^2 + (-35/32)^2]

|z| = sqrt (625/2304 + 1225/1024)

|z| = sqrt (625*4/2304 + 1225*9/1024)

|z| = sqrt (2500+11025)/9216

**|z| = (sqrt 13525) / 96**