3z - 2i = (5+i) / ( 1-i)

First we need to re-write z using the form z =a+ bi

First we will multiply and divide the right side by ( 1+ i)

==> 3z-2i = ( 5+ i)(1+i) / (1-i)(1+i)

==> 3z - 2i = ( 5 + 6i + i^2) / ( 1+ 1)

We know that i^2 = -1

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

==> 3z - 2i = 2 + 3i

Now let us add 2i to both sides:

==> 3z = 2+ 3i + 2i

==> 3z = 2 + 5i

Now divide by 3:

==> z = (2/3) + (5/3) i

Now we will calculate the absolute values:

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

= sqrt( 2/3)^2 + ( 5/3)^2

= sqrt( 4+ 25)/9

- sqrt(29) / 3

**==>The absolute value for z is : l Z l = sqrt29 / 3**

To find the absolute value of z in 3x-2i = (5+i)/(1-i).

We have to solve this for z or 3z first in x+yi form and the find the absolute value of 3|z| = |x+yi| = sqrt(x^2+y^2) and |z| = (1/3)sqrt(x^2+y^2).

3z-2i = (5+i)/(1-i) = (5+i)(1+i)/(1-i)(1+i) = (5-1 +5i+i)/2 , as i^2= -1/

Therefore 3z -2i = (4+6i)/2

3z - 2i = 2+3i +2i = 2+5i.

Taking modulous (or absolute value) on both sides we get:

|3z| = 3|z| = |2+5i|

3|z|= sqrt(2^2+5^2) = sqrt29.

3|z| = sqrt29.

|z| = (sqrt29)/3.