Problem 15 - Precalculus

Question

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

Step-by-Step Solution

Answer

The expression evaluates to \(30 + 19i\).

Step 1: Write the Expression

We need to evaluate the expression for the product of two complex numbers: \(z = 2 + 3i\) and \(w = 9 - 4i\). First, write down the expression \(zw\), which means \((2 + 3i)(9 - 4i)\).

Step 2: Apply the Distributive Property

To multiply the complex numbers, apply the distributive property (similar to FOIL for binomials): \((2 + 3i)(9 - 4i)\) = \(2 \times 9 + 2 \times (-4i) + 3i \times 9 + 3i \times (-4i)\).

Step 3: Compute Each Term

Now compute the individual products: 1. \(2 \times 9 = 18\)2. \(2 \times -4i = -8i\)3. \(3i \times 9 = 27i\)4. \(3i \times -4i = -12i^2\).

Step 4: Simplify the Complex Product

Combine the terms: \(18 - 8i + 27i - 12i^2\). Notice that \(i^2 = -1\), so \(-12i^2 = 12\) because multiplying by \(-1\) changes the sign.

Step 5: Combine Like Terms

Combine the real and imaginary parts: Real: \(18 + 12 = 30\)Imaginary: \(-8i + 27i = 19i\).So, \(zw = 30 + 19i\).

Key Concepts

Distributive Property

When multiplying two complex numbers, we use a method similar to the FOIL method for binomials. This approach is known as the distributive property, which allows us to multiply each term inside one complex number by each term within another.
In simpler terms, if you have an expression such as

• (2 + 3i)(9 - 4i),
• first multiply 2 by both terms in the second complex number: 2 × 9 and 2 × -4i.
• Then, multiply 3i by each term in the second complex number: 3i × 9 and 3i × -4i.

This breaks down larger multiplication problems into smaller, more manageable steps. After computing these small products, the next step is to gather the results together into a single expression. By expanding out the terms, we have all we need to simplify and solve the multiplication of complex numbers.
Just like in the original exercise, distributing the terms correctly allows you to handle complex outputs properly such as in:

• 18 and 12 (real parts of the final answer)
• -8i and 27i (imaginary parts that are combined later).

Imaginary Unit

The imaginary unit, denoted as \(i\), is a fascinating part of mathematics. It is defined as the square root of -1, meaning \(i^2 = -1\). Complex numbers, such as those you saw in the exercise, include this imaginary unit to accommodate calculations that standard real numbers cannot handle.
Here's how it works in the context of multiplication: when you multiply two imaginary numbers or deal with terms that involve \(i\), you may often encounter \(i^2\). For example, in our calculation:

• 3i × -4i results in -12i^2.

To handle such terms properly, replace \(i^2\) with -1. So,

• -12i^2 becomes 12 because -12 multiplied by -1 equals 12.

By understanding this fundamental property of \(i\), you can easily simplify complex expressions, turning them into more familiar real numbers. As we saw:

• the combination of 18 and 12 led to 30 as the real part,
• while the imaginary portion became 19i by balancing the imaginary terms.

Knowing how to manipulate \(i\) ensures you can handle complex numbers deftly in mathematical expressions.

Complex Conjugate

Though not directly used in the multiplication done in the exercise, the concept of a complex conjugate is vital in dealing with complex numbers. A complex conjugate involves flipping the sign of the imaginary part of a complex number.
For example, the complex conjugate of 9 - 4i is 9 + 4i. This transformation is easy to perform: simply change the sign in front of the \(i\).
Complex conjugates are exceptionally useful because they help simplify division involving complex numbers and have other applications, such as finding the magnitude of a complex number. Multiplying a complex number by its conjugate results in a real number. If you took a complex number like \(z = 2 + 3i\) and its conjugate, \(2 - 3i\), their product results in a non-complex, real number.

• (2 + 3i)(2 - 3i) = 4 - 9i^2
• Since \(i^2 = -1\), the equation becomes 4 + 9 = 13, a real number.

This ability to eliminate the imaginary component makes complex conjugates a powerful tool in complex arithmetic. Once you familiarize yourself with how conjugates work, you'll unlock yet more of what makes complex numbers so useful and versatile in mathematics.