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

( 2-4i)/ (1-i) + (3+i)/2

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

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

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

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

=> (9 - i )/2

**Therefore z = ( 2-4i)/ (1-i) + (3+i)/2 = (9-i)/2**

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

We need to simplify the number by writing the number into the complex number form "z = a+ bi".

First we will simplify the first part.

==> (2-4i)/(1-i)

We will multiply and divide by the denominator inverse ( 1+i)

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

= (6-2i)/2 = **3-i....(1)**

Now we will simplify the second part:

(3+i)/2 = **(3/2) + (1/2)i........(2)**

Now we'll add (1) and (2)

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

==> Z = (3 + 3/2 ) + (-1 + 1/2) i

**==> z = (9/2) - (1/2)i**