z= (2-3i) / (2+i)

To find the absolute value for a complex number, first we need to rewrite in the standard form z = a+ bi

We will multiply both numerator and denominator by the inverse of the denominator:

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

Let us simplify the numerator first:

(2-3i)(2-i) = 4 - 2i - 6i +3i^2

but i^2 = -1

==> (2-3i)(2-i) = 4 - 8i - 3= 1- 8i

Now we will simplify the denominator:

(2+i)(2-i) = (4 - i^2 = 4+1 = 5

==> z = (1-8i)/5

==> z = 1/5 - (8/5)i

==> l z l = sqrt(a^2 + b^2 )

= sqrt(1/25 + 64/25) = sqrt65 / 5

z =* (2-3i)/(2+i)*

let z = a + bi

(a + bi )(2 + i) = 2 - 3i

2a + 2bi + ai - b = 2 - 3i

This creates a system of two equations by combining real and imaginary components:

2a - b = 2

2b + a = -3

Solving,

a = 1/5, b = -8/5

So, * (2-3i)/(2+i) *= 0.2 - 1.6i

r = 1.6 ,theta = -82.9° in polar coordinates.

Thus the absolute value is 1.6