The problem asks to simplify the expression involving complex numbers: $0.5i^{12} + \frac{(3i+1)(i-3)}{i-1}$

1. Problem Description

The problem asks to simplify the expression involving complex numbers:

$0.5i^{12} + \frac{(3i+1)(i-3)}{i-1}$

2. Solution Steps

First, let's simplify $i^{12}$. We know that

$i^2 = -1$, $i^3 = -i$, and $i^4 = 1$.

Thus, $i^{12} = (i^4)^3 = 1^3 = 1$.

Therefore, $0.5 i^{12} = 0.5 \times 1 = 0.5$.

Next, simplify the numerator $(3i+1)(i-3)$:

$(3i+1)(i-3) = 3i^2 - 9i + i - 3 = 3(-1) - 8i - 3 = -3 - 8i - 3 = -6 - 8i$.

So, the expression becomes:

$0.5 + \frac{-6 - 8i}{i-1}$.

To divide complex numbers, multiply the numerator and denominator by the conjugate of the denominator.

The conjugate of $i-1$ is $-i-1$.

$\frac{-6-8i}{i-1} = \frac{(-6-8i)(-i-1)}{(i-1)(-i-1)} = \frac{6i+6+8i^2+8i}{-i^2 - i + i + 1} = \frac{14i+6+8(-1)}{-(-1)+1} = \frac{14i+6-8}{1+1} = \frac{14i-2}{2} = 7i-1$.

Therefore,

$0.5 + \frac{-6 - 8i}{i-1} = 0.5 + (7i-1) = 0.5 + 7i - 1 = -0.5 + 7i$.

3. Final Answer

The final answer is $-0.5 + 7i$.