Problem 21 - Precalculus

Question

Evaluate each expression using the values \(z=2+3 i, w=9-4 i,\) and \(w_{1}=-7-i\). $$z\left(w+w_{1}\right)$$

Step-by-Step Solution

Answer

19 - 4i

Step 1: Substitute the Values

First, substitute the given values for \(z\), \(w\), and \(w_1\) into the expression \(z(w+w_1)\). This gives us: \((2+3i)((9-4i) + (-7-i))\).

Step 2: Simplify Inside the Parenthesis

Simplify the expression inside the parenthesis by combining like terms. \((9-4i) + (-7-i) = (9 - 7) + (-4i - i) = 2 - 5i\). Now, our expression becomes \((2+3i)(2-5i)\).

Step 3: Distribute z

Distribute \(z = 2+3i\) across \(2-5i\) using the distributive property. This can be done by multiplying each term in \(2+3i\) by each term in \(2-5i\).

Step 4: Multiply the Real and Imaginary Parts

Calculate: - \(2 imes 2 = 4\)- \(2 imes (-5i) = -10i\)- \(3i imes 2 = 6i\)- \(3i imes (-5i) = -15i^2\)

Step 5: Simplify the Result

Combine like terms:- Combine the real parts: \(4 + 15 = 19\) (since \(i^2 = -1\), so \(-15i^2 = 15\))- Combine the imaginary parts: \(-10i + 6i = -4i\)So the expression becomes \(19 - 4i\).

Key Concepts

Complex Number Arithmetic

Complex number arithmetic involves performing operations such as addition, subtraction, multiplication, and division on numbers of the form \(a+bi\). Here, \(a\) is the real part, and \(bi\) is the imaginary part.
When working with complex numbers, the key is to treat them much like polynomials. You can apply the rules of arithmetic while being mindful of the imaginary unit \(i\), where \(i^2 = -1\).

• Addition and Subtraction: Combine like terms, those with and without the imaginary part, separately.
• Multiplication: Use the distributive property, as you've seen, to simplify expression involving complex numbers.

By mastering these basic operations, you gain the tools needed to solve a variety of mathematical problems involving complex numbers.

Imaginary Numbers

Imaginary numbers introduce the concept of the square root of negative one, represented by the symbol \(i\). When you square \(i\), you get \(-1\). This property sets the foundation for many intriguing results in mathematics.
Real numbers alone cannot provide solutions to certain equations, like \(x^2 + 1 = 0\). To solve these, mathematicians introduced imaginary numbers:

• Form: Imaginary numbers typically appear as \(bi\), where \(b\) is a real number.
• Use: They allow for the expression and solution of equations that would otherwise lack real solutions.

Understanding imaginary numbers can seem abstract at first, but they become invaluable in complex number arithmetic and many fields of mathematics.

Distributive Property

The distributive property is a fundamental principle in algebra that allows you to multiply a single term across an expression inside a parenthesis. It's expressed as \(a(b + c) = ab + ac\) for real numbers.
In complex number arithmetic, you apply this property similarly:

• Application: When multiplying two complex numbers, such as \((2+3i)\times(2-5i)\), distribute each term from the first set of parentheses across each term in the second set.
• Process: You perform multiplications like \(2 \times 2\), \(2 \times (-5i)\), \(3i \times 2\), and \(3i \times (-5i)\), as shown in the step-by-step solution.

By applying the distributive property, you systematically unfold complex expressions, making them more manageable and easier to simplify.

Complex Multiplication

Complex multiplication extends the distributive property to include operations involving imaginary numbers. To multiply complex numbers like \((2+3i)\) and \((2-5i)\), you handle the real and imaginary units separately at first but eventually mix them:

• Real and Imaginary Parts: Multiply each pair of terms individually and pay close attention to terms involving \(i\).
• Multiply by \(i\): Remember, \(i^2 = -1\), so expressions like \(-15i^2\) turn into positive real numbers (here, \(15\)).
• Combine: Finally, sum all resulting terms—grouping real parts and imaginary parts separately—to form a single complex number.

With practice, these steps become a routine process, allowing you to handle any complex multiplication with confidence and precision.