Question

the number $4 + i$ is multiplied by some real number to yield $2 + \frac{1}{2}i$. which equation is true?
$\frac{1}{2}(4 + i) = 2 + \frac{1}{2}i$
$2(4 + i) = 2 + \frac{1}{2}i$
$-2(4 + i) = 2 + \frac{1}{2}i$
$-\frac{1}{2}(4 + i) = 2 + \frac{1}{2}i$

Explanation:

Step1: Define the real multiplier

Let the real number be $k$. The equation is $k(4+i) = 2+\frac{1}{2}i$.

Step2: Solve for $k$

Divide both sides by $4+i$:
$k = \frac{2+\frac{1}{2}i}{4+i}$
Multiply numerator and denominator by $4-i$ (the conjugate) to rationalize:
$k = \frac{(2+\frac{1}{2}i)(4-i)}{(4+i)(4-i)}$
Calculate numerator: $(2)(4) + 2(-i) + \frac{1}{2}i(4) + \frac{1}{2}i(-i) = 8 - 2i + 2i - \frac{1}{2}i^2 = 8 + \frac{1}{2} = \frac{17}{2}$
Calculate denominator: $4^2 - i^2 = 16 + 1 = 17$
So $k = \frac{\frac{17}{2}}{17} = \frac{1}{2}$

Step3: Verify the correct equation

Substitute $k=\frac{1}{2}$ into the original form: $\frac{1}{2}(4+i) = 2+\frac{1}{2}i$

Answer:

$\frac{1}{2}(4+i) = 2+\frac{1}{2}i$