Find the absolute value of z if 3z-2i = (5+i)/ (1-i)

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