Question

Find the real solution \((s),\) if any, of each equation. Multiply \((7+3 i)(1-2 i) .\) Write the answer in the form \(a+b i\).

Short answer

13 - 11i

Step-by-step solution

- Use the Distributive Property

To multiply the complex numbers \((7 + 3i)(1 - 2i)\), distribute each term in the first complex number by each term in the second complex number.

- Apply the FOIL Method

First, multiply the first terms: \(7 \times 1 = 7.\) Next, multiply the outside terms: \(7 \times -2i = -14i.\) Then, multiply the inside terms: \(3i \times 1 = 3i.\) Finally, multiply the last terms: \(3i \times -2i = -6i^2.\)

- Combine Like Terms

Combine all the terms you found in the previous step: \(7 - 14i + 3i - 6i^2.\)

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

Recognize that \(i^2 = -1\), so \(-6i^2 = -6(-1) = 6.\) This gives us \(7 - 14i + 3i + 6.\)

- Combine Real and Imaginary Parts

Combine the real parts and the imaginary parts: \(7 + 6 = 13\) and \(-14i + 3i = -11i.\)

- Write the Final Answer

The final answer in the form \(a + bi\) is \13 - 11i.\

Key concepts

Distributive Property

Multiplying complex numbers often involves using the **distributive property** similar to how we multiply polynomials. The distributive property tells us to multiply each term in the first complex number by each term in the second complex number.
For example, given \( (7 + 3i)(1 - 2i) \), each term in *7 + 3i* must be multiplied by each term in *1 - 2i*.

Here's how it works step-by-step:
1. Distribute *7* across \( 1 - 2i \)
* 7 × 1 = 7
* 7 × -2i = -14i

2. Distribute *3i* across \( 1 - 2i \)
* 3i × 1 = 3i
* 3i × -2i = -6i².

Now we've distributed both terms, resulting in four separate products that we need to combine to simplify further.

FOIL Method

The **FOIL method** is a specific application of the distributive property for binomials, and it's particularly helpful with complex numbers. FOIL stands for First, Outside, Inside, and Last terms.

Here’s how to apply it to our example \( (7 + 3i)(1 - 2i) \):

• **First:** Multiply the first terms in each binomial: 7 × 1 = 7.
• **Outside:** Multiply the outermost terms: 7 × -2i = -14i.
• **Inside:** Multiply the inner terms: 3i × 1 = 3i.
• **Last:** Multiply the last terms in each binomial: 3i × -2i = -6i².

After following the FOIL steps, combine all parts to get:

\( 7 -14i + 3i - 6i² \).

Remember, multiplication of imaginary units also plays a critical role in the final step.

Imaginary Unit i

In the context of complex numbers, the **imaginary unit i** is essential. By definition, \( i \) is the square root of -1:

\[ i = \sqrt{-1} \].

The square of \( i \) produces a fundamental property that simplifies many problems:

\[ i^2 = -1 \].

Applying this property, we simplify terms involving \( i^2 \). In our exercise, we found that:

\(-6i^2 = -6(-1) = 6 \.

So, our equation becomes \ 7 - 14i + 3i + 6 \).

Combining like terms, the final result is:
\ 13 - 11i \.

This entire process shows how using the imaginary unit’s properties helps simplify and solve expressions involving complex numbers.