Question

which complex number is equivalent to this expression?
\\(\frac{1}{3}(6 + 3i) - \frac{2}{3}(6 - 12i)\\)
\\(\bigcirc\\) a. \\(6 + 9i\\)
\\(\bigcirc\\) b. \\(-2 - 7i\\)
\\(\bigcirc\\) c. \\(-2 + 9i\\)
\\(\bigcirc\\) d. \\(6 - 7i\\)

Explanation:

Step1: Distribute the fractions

First, distribute \(\frac{1}{3}\) to \(6 + 3i\) and \(\frac{2}{3}\) to \(6 - 12i\).
For \(\frac{1}{3}(6 + 3i)\), we have \(\frac{1}{3} \times 6 + \frac{1}{3} \times 3i = 2 + i\).
For \(\frac{2}{3}(6 - 12i)\), we have \(\frac{2}{3} \times 6 - \frac{2}{3} \times 12i = 4 - 8i\).
So the expression becomes \((2 + i) - (4 - 8i)\).

Step2: Subtract the complex numbers

Subtract the real parts and the imaginary parts separately.
Real part: \(2 - 4 = -2\).
Imaginary part: \(i - (-8i) = i + 8i = 9i\).
Combining them, we get \(-2 + 9i\).

Answer:

C. -2 + 9i