Given that `I = sqrt-1,` what is the product of `(2-4i+ 2i^(2))/(2)and (1)/(1-i)` ?

Options

A. `-1+i`
B. `-1-i`
C. `1-i`
D. `1+i`

Answer

To solve the problem, we need to find the product of the two expressions given:

1. \((2 - 4i + 2i^2) / 2\)
2. \(1 / (1 - i)\)

Let's break it down step by step.

### Step 1: Simplify the first expression \((2 - 4i + 2i^2) / 2\)

We know that \(i^2 = -1\). Therefore, we can substitute \(i^2\) in the expression:

\[ 2 - 4i + 2i^2 = 2 - 4i + 2(-1) = 2 - 4i - 2 \]

This simplifies to:

\[ 0 - 4i = -4i \]

Now, we divide by 2:

\[ \frac{-4i}{2} = -2i \]

### Step 2: Simplify the second expression \(1 / (1 - i)\)

To simplify \(1 / (1 - i)\), we will rationalize the denominator. We multiply the numerator and the denominator by the conjugate of the denominator, which is \(1 + i\):

\[ \frac{1}{1 - i} \cdot \frac{1 + i}{1 + i} = \frac{1 + i}{(1 - i)(1 + i)} \]

Now, calculate the denominator:

\[ (1 - i)(1 + i) = 1^2 - i^2 = 1 - (-1) = 1 + 1 = 2 \]

So we have:

\[ \frac{1 + i}{2} \]

### Step 3: Multiply the results from Step 1 and Step 2

Now we multiply the results from Step 1 and Step 2:

\[ -2i \cdot \frac{1 + i}{2} \]

This simplifies to:

\[ \frac{-2i(1 + i)}{2} = -i(1 + i) \]

Expanding this gives:

\[ -i - i^2 \]

Since \(i^2 = -1\), we substitute:

\[ -i - (-1) = -i + 1 = 1 - i \]

### Final Answer

Thus, the product of the two expressions is:

\[ 1 - i \]