Problem 29 - Algebra and Trigonometry with Analytic Geometry

Question

Write the expression in the form \(a+b i\), where \(a\) and \(b\) are real numbers. $$ (2-\sqrt{-4})(3-\sqrt{-16}) $$

Step-by-Step Solution

Answer

The expression in the form \( a + bi \) is \(-2 - 14i \).

Step 1: Identify the Imaginary Components

Recognize that the expression \(\sqrt{-4}\) and \(\sqrt{-16}\) involve the imaginary unit \(i\). We have: \(\sqrt{-4} = 2i\) and \(\sqrt{-16} = 4i\).

Step 2: Substitute the Imaginary Units

Substitute the imaginary parts in the expression with their respective values. The expression becomes: \( (2 - 2i)(3 - 4i) \).

Step 3: Expand the Expression

Use the distributive property, (FOIL Method), to expand:- First: \(2 \times 3 = 6\)- Outer: \(2 \times (-4i) = -8i\)- Inner: \((-2i) \times 3 = -6i\)- Last: \((-2i) \times (-4i) = 8i^2\) Combine all the terms to get: \(6 - 8i - 6i + 8i^2\).

Step 4: Simplify the Expression

Since \(i^2 = -1\), replace \(8i^2\) with \(-8\):\(6 - 8i - 6i + (-8) = 6 - 14i - 8\).Combine like terms to get \(-2 - 14i\).

Step 5: Write in the Form a+bi

The expression \( -2 - 14i \) is already in the form \( a + bi \) where \( a = -2 \) and \( b = -14 \).

Key Concepts

Imaginary Unit

In the realm of mathematics, the imaginary unit is denoted by the letter \(i\). This unit is crucial in complex numbers. It is defined as \(i = \sqrt{-1}\).
This is a fundamental concept, as it allows us to express the square root of negative numbers, which is impossible within the real number system.
In our exercise, we encounter \(\sqrt{-4}\) and \(\sqrt{-16}\). Here's how we solve them using the imaginary unit:

• \(\sqrt{-4} = \sqrt{4} \cdot \sqrt{-1} = 2i\)
• \(\sqrt{-16} = \sqrt{16} \cdot \sqrt{-1} = 4i\)

By substituting these values back into our expression, we convert a seemingly complex task into an easier one by viewing them as complex numbers.

Distributive Property

The distributive property is a useful algebraic rule for solving problems involving multiplications over additions or subtractions.
This property lets us handle expressions like \((a - bi)(c - di)\) by expanding it step by step according to specific rules.
If we want to expand two binomials, we apply what's known as the FOIL method.- **First terms**: Multiply the first terms of each binomial, \(2 \times 3 = 6\).- **Outer terms**: Multiply the outer terms, \(2 \times (-4i) = -8i\).- **Inner terms**: Multiply the inside terms, \((-2i) \times 3 = -6i\).- **Last terms**: Multiply the last terms, \((-2i) \times (-4i) = 8i^2\).This results in the expression \(6 - 8i - 6i + 8i^2\), merging beautifully due to the distributive approach.

Complex Conjugate

In complex arithmetic, a complex conjugate is a valuable concept when working with complex numbers.
The complex conjugate of a number \(a + bi\) is \(a - bi\), flipping the sign of the imaginary part.
Although the exercise doesn't explicitly ask for complex conjugates, recognizing this concept is crucial.By using conjugates, we can:

• Simplify division of complex numbers by clearing the imaginary component from the denominator.
• Ensure pure real results when needed, as any complex number times its conjugate is always real.

Understanding complex conjugates prepares us for more advanced topics like polar forms and power calculations in complex arithmetic.

Complex Arithmetic

Complex arithmetic extends the basic arithmetic operations to complex numbers, which include both real and imaginary parts.
This involves addition, subtraction, multiplication, and division of complex numbers.
In the exercise, the steps involve substituting, distributing, and eventually simplifying a complex expression into its standard form \(a + bi\).After expansion and application of foundational properties, we find:

• Replacement of \(i^2\) using the relation \(i^2 = -1\).
• Simplification of terms: \(6 - 8i - 6i + 8i^2 \to 6 - 14i - 8\).
• Final result is \(-2 - 14i\) as \(a + bi\).

Grasping complex arithmetic allows one to solve a broad range of problems involving complex numbers efficiently.