Solve the equation 2iz/(1 - i) + 2i = 2z(1 - i) - 3 + i

We'll note z = a + b*i and we'll substitute z into the
given equation.

2i*(a + b*i)/(1-i) + 2i = 2(a + b*i)(1-i) -
3 + i

We'll multiply the terms 2i,  2(a + b*i)(1-i),  - 3
and i by (1-i):

2i*(a + b*i) + 2i*(1-i) = 2(a + b*i)(1-i)^2
- 3(1-i) + i*(1-i)

We'll remove the
brackets:

2ai - 2b + 2i + 2 = -4ai + 4b - 3 + 3i + i +
1

We'll move all the terms in a and b to the left side and
all the terms without a and b, to the right side:

2ai - 2b
+ 4ai - 4b = -2i - 2 - 3 + 3i + i + 1

We'll combine the
real parts and the imaginary parts:

-6b + 6ai = -4 +
2i

Because the expressions from both sides are equivalent,
the real parts and the imaginary parts have to be
equal.

-6b =
-4

b =
2/3

6a =
2

a =
1/3

z = 1/3 +
2i/3