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**

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**

To simplify the complex number z = (2-4i)/(1-i) +(3+i)/2.

We bring the given complex number to the form x+yi, where x and y are real.

First term = (2-4i)/(1-i) = (2-4i)(1+i)/(1-i)(1+i). We multiplied numerator and denominator by 1+i.

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

(2-4i)/(1-i) = (2-2i+4)/(1+1) , as i^2 = -1

(2-4i)/(1-i) = (6-2i)/2

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

2nd term = (3+i)/2 = 3/2 +i/2.

adding the two terms , we get: (3-i) +(3/2+i/2) = (3+3/2) +(-i+i/2) = 9/2 -i/2

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

Therefore the simplified form of complex number = 4.5 -0.5i.