Problem 44 - Precalculus Mathematics for Calculus

Question

Evaluate the expression and write the result in the form \(a+b i\) $$\frac{-3+5 i}{15 i}$$

Step-by-Step Solution

Answer

The expression simplifies to \( \frac{1}{3} + \frac{1}{5}i \).

Step 1: Simplify the Denominator

To simplify the expression \( \frac{-3+5i}{15i} \), start by expressing the complex denominator in a simpler form. We can multiply both the numerator and denominator by \( i \) to make the denominator real. So the denominator becomes: \(15i \cdot i = 15i^2 = 15(-1) = -15\).

Step 2: Simplify the Numerator

Let’s simplify the numerator by multiplying \(-3 + 5i\) by \( i \) as well: \((-3 + 5i) \cdot i = -3i + 5i^2 = -3i + 5(-1) = -3i - 5.\)Thus, the new numerator is \(-5 - 3i\).

Step 3: Divide Numerator by Denominator

Now divide the simplified numerator by the simplified denominator:\(\frac{-5 - 3i}{-15} = \frac{-5}{-15} + \frac{-3i}{-15}.\)This breaks down into two separate fractions which can be simplified individually.

Step 4: Simplify Each Fraction

Simplify each fraction:\( \frac{-5}{-15} = \frac{5}{15} = \frac{1}{3}, \) and \(\frac{-3i}{-15} = \frac{3i}{15} = \frac{i}{5}.\)Thus, the expression simplifies to \( \frac{1}{3} + \frac{1}{5}i. \)

Step 5: Final Answer in Standard Form

Combine the real part and the imaginary part to express the answer in the standard form \(a + bi\):The expression \( \frac{1}{3} + \frac{1}{5}i \) is already in the correct form. Therefore, if you need to express it as a common denominator, it stays as \( \frac{1}{3} + \frac{1}{5}i \).

Key Concepts

Imaginary Unit

In mathematics, not all numbers can be expressed as real numbers. Sometimes, we encounter situations that require an imaginary unit, often denoted as \( i \). By definition, \( i \) is the solution to the equation \( x^2 = -1 \). It's essentially a tool that lets us handle negative square roots, which aren't possible in the realm of real numbers.

Since \( i^2 = -1 \), any power of \( i \) can be simplified by using this property:

• \( i^1 = i \)
• \( i^2 = -1 \)
• \( i^3 = -i \)
• \( i^4 = 1 \)

These properties repeat every four powers, forming a cycle. This cyclic behavior is crucial when simplifying expressions, especially when we multiply complex numbers by \( i \). To illustrate, multiplying a complex number by \( i \) rotates it by 90 degrees in the complex plane, a handy concept in both mathematics and engineering.

Complex Conjugate

The complex conjugate of a complex number helps in various simplification processes, especially division involving complex numbers. Given a complex number \( a + bi \), its complex conjugate is \( a - bi \). By changing the sign of the imaginary part, complex conjugates have a unique property: when a complex number is multiplied by its conjugate, the result is a real number.

For example, consider the number \( 3 + 4i \). Its complex conjugate is \( 3 - 4i \). When these are multiplied together, we get:\[(3 + 4i)(3 - 4i) = 9 - 12i + 12i - 16i^2 = 9 + 16 = 25.\]Notice how the imaginary parts cancel out, making the result purely real. This feature is incredibly useful when dividing complex numbers, converting the denominator into a real number and simplifying the expression. While the original exercise didn’t involve explicitly using a complex conjugate, understanding this concept is fundamental for dealing with more complex operations involving complex numbers.

Simplifying Fractions

When working with complex numbers in fractional form, simplifying is key to make the expression more manageable and understandable. Simplifying fractions often involves reducing fractions to their simplest terms and sometimes factoring out common components.

In complex expressions, like our problem's original form \( \frac{-3+5i}{15i} \), simplification begins with tackling both the numerator and the denominator. We strive to eliminate any complex number from the denominator to make it real by multiplying by \( i \). This results in the denominator \( 15i^2 = -15 \), changing our expression to \( \frac{-5 - 3i}{-15} \).

Breaking this down further into two individual fractions, we aim to achieve simpler representations: \( \frac{-5}{-15} \) and \( \frac{-3i}{-15} \). Each of these fractions can then be reduced separately, resulting in a simple form \( \frac{1}{3} + \frac{i}{5} \). This method ensures clarity and correctness, presenting the final expression in a neat, standardized format \( a + bi \).