Given the complex number z = (2i-1)/(3+4i)

We need to find the absolute values of z = l z l.

First we need to simplify z and rewrite into the standard form z= a+ bi.

First we will multiply both numerator and denominator by (3-4i).

==> z = (-1+2i)(3-4i)/(3+4i)(3-4i)

= (-3 +4i +6i -8i^2)/(9 -16i^2)

But we know that i^2 = -1

==> z = ( -3+10i +8)/(9+16)

= ( 5+10i) / 25

==> z = (1/5) + (2/5)i

Now we will calculate the absolute values.

We know that l zl = sqrt(a^2+b^2)

==> l z l = sqrt(1/5)^2 + (2/5)^2

= sqrt(1+4)/25 = sqrt5 / 5

**==> l z l = sqrt5 / 5 = 1/sqrt5**

We have z= (2i-1)/(3+4i).

Now z= (2i-1)/(3+4i)

=> z = ( 2i - 1)*(3 - 4i) / (3 + 4i)(3 - 4i)

=> z = (6i - 3 - 8i^2 + 4i) / (3^2 - (4i)^2)

=> z = (10i - 3 + 8) / (9 + 16)

=> z = (10i + 5)/25

=> z = (10/25)i + (5/25)

Now |z| = sqrt ( (10/25)^2 + (5/25)^2)

=> |z| = sqrt ( 125) / 625)

=> |z| = sqrt 5/ 25

=> |z| = 1/ sqrt 5

**Therefore the absolute value of z is 1/ sqrt 5.**

To find the absolute value of z = (2i-1)/(3+4i).

Let us first find z in the form x+yi and then determine absolute z = sqrt(x^2+y^2).

z = (2i-1)/(3+4i).

We multiply the right numerator and denominator by 3-4i.

z = (2i-1)((3-4i)/(3^2-4^2*i^2) = (2i-1)(3-4i)/25.

z= (1/25)(6i-8i^2-3 +4i).

z = (1/25) { 10i+8-3) as i^2= 1.

z = (1/25){ 5+10i)

z = (5/25){ 1+2i).

z = (1/5)+ (2/5)i.

Absolute z = {(1/5)^2+(2/5)^2 = (1/5)(1+4)^(1/2) = (sqrt5)/5.