Calculate |z| if 8z+40z'=25+35i?

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