We are asked to simplify the complex number expression: $0.5i^{12} + \frac{(3i+1)(i-3)}{i-1}$.

1. Problem Description

We are asked to simplify the complex number expression:

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

2. Solution Steps

First, we simplify $i^{12}$.

We know that $i^2 = -1$, $i^3 = -i$, $i^4 = 1$.

Since $12$ is a multiple of $4$, $i^{12} = (i^4)^3 = 1^3 = 1$.

Therefore, $0.5i^{12} = 0.5(1) = 0.5$.

Next, we expand the numerator of the fraction:

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

So the expression becomes $0.5 + \frac{-6 - 8i}{i-1}$.

To simplify the fraction, we multiply the numerator and denominator by the conjugate of the denominator, which is $-i-1$ or $-1-i$.

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

Finally, we add the two parts together:

$0.5 + (-1 + 7i) = 0.5 - 1 + 7i = -0.5 + 7i$.

3. Final Answer

$-0.5 + 7i$