Question

8 matching 3 points simplify and match each complex number. (2+3i)(4-2i) (2-i)(2+i) (5+i)(2-3i)

Explanation:

Simplifying \((2 + 3i)(4 - 2i)\)

Step 1: Use the distributive property (FOIL method)

Multiply each term in the first complex number by each term in the second complex number:
\[

$$\begin{align*} (2 + 3i)(4 - 2i)&=2\times4+2\times(-2i)+3i\times4+3i\times(-2i)\\ &=8 - 4i + 12i - 6i^2 \end{align*}$$

\]

Step 2: Simplify using \(i^2=-1\)

Combine like terms and substitute \(i^2 = - 1\):
\[

$$\begin{align*} 8 - 4i + 12i - 6i^2&=8 + 8i-6\times(-1)\\ &=8 + 8i + 6\\ &=14 + 8i \end{align*}$$

\]

Simplifying \((2 - i)(2 + i)\)

Step 1: Recognize the difference of squares formula \((a - b)(a + b)=a^2 - b^2\)

Here, \(a = 2\) and \(b = i\), so:
\[
(2 - i)(2 + i)=2^2 - i^2
\]

Step 2: Simplify using \(i^2=-1\)

Substitute \(i^2=-1\) into the expression:
\[

$$\begin{align*} 2^2 - i^2&=4-(-1)\\ &=4 + 1\\ &=5 \end{align*}$$

\]

Simplifying \((5 + i)(2 - 3i)\)

Step 1: Use the distributive property (FOIL method)

Multiply each term in the first complex number by each term in the second complex number:
\[

$$\begin{align*} (5 + i)(2 - 3i)&=5\times2+5\times(-3i)+i\times2+i\times(-3i)\\ &=10-15i + 2i-3i^2 \end{align*}$$

\]

Step 2: Simplify using \(i^2=-1\)

Combine like terms and substitute \(i^2 = - 1\):
\[

$$\begin{align*} 10-15i + 2i-3i^2&=10-13i-3\times(-1)\\ &=10-13i + 3\\ &=13-13i \end{align*}$$

\]

Answer:

s:

• \((2 + 3i)(4 - 2i)=14 + 8i\)
• \((2 - i)(2 + i)=5\)
• \((5 + i)(2 - 3i)=13-13i\)