Question

1. \frac{16 - 32i}{-8i}

Explanation:

Step1: Simplify numerator and denominator

Factor out 16 from numerator: \( \frac{16(1 - 2i)}{-8i} \)
Simplify the fraction: \( \frac{16}{-8} \cdot \frac{1 - 2i}{i} = -2 \cdot \frac{1 - 2i}{i} \)

Step2: Rationalize the denominator

Multiply numerator and denominator by \( i \): \( -2 \cdot \frac{(1 - 2i)i}{i \cdot i} \)
Simplify denominator: \( i^2 = -1 \), so denominator becomes \( -1 \)
Simplify numerator: \( (1 - 2i)i = i - 2i^2 = i + 2 \) (since \( i^2 = -1 \))
Now we have: \( -2 \cdot \frac{2 + i}{-1} \)

Step3: Simplify the expression

\( -2 \cdot (- (2 + i)) = 2(2 + i) = 4 + 2i \)
Wait, let's check again. Wait, maybe a better way:
Original expression: \( \frac{16 - 32i}{-8i} \)
Divide numerator and denominator by -8: \( \frac{-2 + 4i}{i} \)
Now multiply numerator and denominator by \( i \): \( \frac{(-2 + 4i)i}{i^2} = \frac{-2i + 4i^2}{-1} = \frac{-2i - 4}{-1} = 4 + 2i \)
Wait, no, let's do it step by step correctly.
First, \( \frac{16 - 32i}{-8i} = \frac{16}{-8i} - \frac{32i}{-8i} \)
Simplify each term:
\( \frac{16}{-8i} = \frac{-2}{i} \), multiply numerator and denominator by \( i \): \( \frac{-2i}{i^2} = \frac{-2i}{-1} = 2i \)
\( \frac{-32i}{-8i} = 4 \)
So adding them together: \( 4 + 2i \)
Wait, that's correct. So the result is \( 4 + 2i \)
Wait, let's verify:
Multiply \( (4 + 2i) \) by \( -8i \): \( 4(-8i) + 2i(-8i) = -32i -16i^2 = -32i +16 \) (since \( i^2 = -1 \)), which is \( 16 -32i \), matching the numerator. So correct.

Answer:

\( 4 + 2i \)