The imaginary number "I" is such that `i^(2)=-1`. Which of the following statements is true about the complex number equivalent to `(4-i)xx(1+2i)+(1-i)xx(2-3i)`?

Options

A. It lies on the real axis
B. It lies in the first quadrant
C. It lies in the second quadrant
D. It lies in the third quadrant

Answer

To solve the expression \((4 - i)(1 + 2i) + (1 - i)(2 - 3i)\), we will follow these steps:

### Step 1: Expand the first product \((4 - i)(1 + 2i)\)

Using the distributive property (also known as the FOIL method for binomials):

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

Calculating each term:

\[ = 4 + 8i - i - 2i^2 \]

### Step 2: Substitute \(i^2 = -1\)

Since \(i^2 = -1\), we can replace \(-2i^2\) with \(2\):

\[ = 4 + 8i - i + 2 \]

### Step 3: Combine like terms

Now, combine the real parts and the imaginary parts:

\[ = (4 + 2) + (8i - i) = 6 + 7i \]

### Step 4: Expand the second product \((1 - i)(2 - 3i)\)

Using the distributive property again:

\[ (1 - i)(2 - 3i) = 1 \cdot 2 + 1 \cdot (-3i) - i \cdot 2 - i \cdot (-3i) \]

Calculating each term:

\[ = 2 - 3i - 2i + 3i^2 \]

### Step 5: Substitute \(i^2 = -1\)

Replace \(3i^2\) with \(-3\):

\[ = 2 - 3i - 2i - 3 \]

### Step 6: Combine like terms

Now, combine the real parts and the imaginary parts:

\[ = (2 - 3) + (-3i - 2i) = -1 - 5i \]

### Step 7: Add the results from Step 3 and Step 6

Now, we add the two results together:

\[ (6 + 7i) + (-1 - 5i) = (6 - 1) + (7i - 5i) = 5 + 2i \]

### Final Result

The complex number equivalent to the expression is:

\[ 5 + 2i \]

### Step 8: Determine the quadrant

Since the real part \(5\) is positive and the imaginary part \(2\) is also positive, the complex number \(5 + 2i\) lies in the first quadrant.