Question

multiply the following:
$(4 + 2i)(-1 + i)$
$\circ -6 + 2i$
$\circ 2 + 2i$
$\circ -4 + 2i + 2i^2$
$\circ -6 + 6i$

Explanation:

Step1: Apply distributive property (FOIL)

Multiply each term in the first complex number by each term in the second complex number:
$$(4 + 2i)(-1 + i)=4\times(-1)+4\times i+2i\times(-1)+2i\times i$$

Step2: Simplify each product

Calculate each product:
$$4\times(-1)= -4$$
$$4\times i = 4i$$
$$2i\times(-1)= -2i$$
$$2i\times i = 2i^{2}$$
So now we have:
$$-4 + 4i-2i + 2i^{2}$$

Step3: Combine like terms and use \(i^{2}=-1\)

Combine the \(i\) terms: \(4i-2i = 2i\)
Substitute \(i^{2}=-1\) into \(2i^{2}\): \(2i^{2}=2\times(-1)= -2\)
Now we have:
$$-4 + 2i-2$$
Combine the constant terms: \(-4-2=-6\)
So the result is:
$$-6 + 2i$$

Answer:

\(-6 + 2i\) (corresponding to the first option: \(-6 + 2i\))