Problem 36 - Algebra 2 and Trigonometry

Question

In \(26-37,\) find each product. $$ \left(\frac{1}{5}-\frac{1}{10} i\right)(4+2 i) $$

Step-by-Step Solution

Answer

The product is 1.

Step 1: Expand the Expression

To find the product of two complex numbers, use the distributive property. The expression is \( (\frac{1}{5} - \frac{1}{10}i)(4 + 2i) \). Expand by multiplying each part: \( \frac{1}{5} \times 4 \), \( \frac{1}{5} \times 2i \), \( -\frac{1}{10}i \times 4 \), and \( -\frac{1}{10}i \times 2i \).

Step 2: Calculate Real Parts

Calculate the real parts of the expressions: \( \frac{1}{5} \times 4 = \frac{4}{5} \) and \( -\frac{1}{10}i \times 2i \). Remember that \( i^2 = -1 \), so \( -\frac{1}{10}i \times 2i = \frac{-2}{10}(-1) = \frac{2}{10} = \frac{1}{5} \). Add these results together to find the real part: \( \frac{4}{5} + \frac{1}{5} = 1 \).

Step 3: Calculate Imaginary Parts

Calculate the imaginary parts of the expressions: \( \frac{1}{5} \times 2i = \frac{2}{5}i \) and \( -\frac{1}{10}i \times 4 = -\frac{4}{10}i = -\frac{2}{5}i \). Add these two results together to find the imaginary part: \( \frac{2}{5}i - \frac{2}{5}i = 0 \).

Step 4: Combine the Real and Imaginary Parts

Combine the real and imaginary parts to form the final product. The real part is 1, and the imaginary part is 0. Therefore, the product is \( 1 + 0i = 1 \).

Key Concepts

Distributive Property

When multiplying complex numbers, the distributive property plays a crucial role. Just like with real numbers, this property simplifies expressions by allowing us to multiply each term in one group by every term in another:

• First, multiply each part of the first expression by each part of the second expression.
• Essentially, you apply the formula: \((a + bi)(c + di) = ac + adi + bci + bdi^2\).

The expansion here respects each component separately, facilitating the handling of complex numbers. By distributing, you ensure that each component interacts correctly with the others, preserving the intended combination of real and imaginary parts.
In the given exercise, by applying the distributive property in \((\frac{1}{5} - \frac{1}{10}i)(4 + 2i)\), we perform these calculations:

• \(\frac{1}{5} \times 4\)
• \(\frac{1}{5} \times 2i\)
• \(-\frac{1}{10}i \times 4\)
• \(-\frac{1}{10}i \times 2i\)

Each multiplication step neatly isolates terms, making subsequent calculations more straightforward. This handling ultimately aligns terms for efficient addition.

Real and Imaginary Parts

Understanding real and imaginary parts is fundamental when dealing with complex numbers. Each complex number consists of:

• A real part, which appears without the imaginary unit \(i\).
• An imaginary part, which is accompanied by the \(i\).

For any complex number expressed as \(a + bi\):

• \(a\) is the real part.
• \(bi\) is the imaginary part.

In our exercise, the products yield numbers that need to be grouped into these categories:
The real components are derived from:

• \(\frac{1}{5} \times 4 = \frac{4}{5}\)
• \(-\frac{1}{10}i \times 2i = \frac{1}{5}\) (after considering \(i^2 = -1\))

These real parts are added to form a sum of \(1\).
The imaginary parts are determined from:

• \(\frac{2}{5}i\) and \(-\frac{2}{5}i\)

Adding them results in zero. Both parts are then combined to express the final result as \(1 + 0i\).

i Squared Equals Negative One

One unique aspect of working with imaginary numbers is the property that \(i^2 = -1\). This relation arises from the definition of imaginary units and is vital to simplifying expressions involving complex numbers.

• The imaginary unit \(i\) is defined such that \(i = \sqrt{-1}\).
• Therefore, squaring both sides gives \(i^2 = -1\).

This property is crucial in the given solution because during the expansion, terms like \(-\frac{1}{10}i \times 2i\) arise. Here:

• The multiplication goes as: \(-\frac{1}{10} \times 2 \times (i \times i)\) simplifies to \(\frac{-2}{10} \times -1 = \frac{2}{10}\).
• This process converts the initially complex expression into a manageable real number.

Remembering that \(i^2\) converts to a real \(-1\) enables the simplification of complex multiplications into more straightforward real numbers, ultimately supporting correct calculations of the sum of real numbers in complex products.