Problem 19 - Advanced Engineering Mathematics

Question

$$\frac{2-11 i}{6-i} \cdot \frac{6+i}{6+i}=\frac{23-64 i}{37}=\frac{23}{37}-\frac{64}{37} i$$

Step-by-Step Solution

Answer

\( \frac{23}{37} - \frac{64}{37} i \) is the simplified result.

Step 1: Express the Problem

We are given the expression \( \frac{2-11i}{6-i} \cdot \frac{6+i}{6+i} \). The goal is to simplify this expression by multiplying the numerator and the denominator of the fraction by the complex conjugate of the denominator.

Step 2: Multiply by the Conjugate

To eliminate the imaginary part from the denominator \( 6-i \), we multiply both the numerator and the denominator by its conjugate \( 6+i \). This gives us: \[ \frac{(2-11i)(6+i)}{(6-i)(6+i)}. \]

Step 3: Simplify the Denominator

The denominator \( (6-i)(6+i) \) is a difference of squares and simplifies to: \[ 6^2 - (i)^2 = 36 + 1 = 37. \]

Step 4: Expand the Numerator

Expand \((2-11i)(6+i)\):\[ 2 \cdot 6 + 2 \cdot i - 11i \cdot 6 - 11i^2 \]= \[ 12 + 2i - 66i + 11 \]= \[ 23 - 64i. \]

Step 5: Final Result

Combine the numerator and the simplified denominator, the expression becomes:\[ \frac{23 - 64i}{37}, \]which can be written as \( \frac{23}{37} - \frac{64}{37} i. \) This matches the given answer.

Key Concepts

Complex Conjugate

In the realm of complex numbers, a complex conjugate is crucial for simplifying expressions. If you have a complex number, say \( a + bi \), its conjugate is \( a - bi \). By changing the sign of the imaginary part, we achieve the complex conjugate. This operation often helps in making calculations more manageable, especially when dealing with fractions involving complex numbers.
For instance, in the given problem, the denominator \( 6 - i \) has its complex conjugate \( 6 + i \). By multiplying the numerator and the denominator by this conjugate, the imaginary part in the denominator disappears, simplifying the expression.
Here’s how it helps:

• Eliminates the imaginary unit in the denominator, transforming it into a real number.
• Allows for easier computation and simplification of the expression.

By doing this transformation, the calculation becomes more straightforward, resulting in easy simplification of complex expressions.

Imaginary Unit

The imaginary unit, represented by \( i \), is a fundamental concept in complex numbers. It is defined as \( i^2 = -1 \). This definition allows us to extend the real numbers into the complex plane.
When we multiply complex numbers, the imaginary unit's property often plays a pivotal role. In our example, while expanding the numerator \( (2 - 11i)(6 + i) \), we face terms like \( -11i^2 \). Because \( i^2 = -1 \), this term becomes \( +11 \), contributing to the real part of the result.
Working with \( i \) requires:

• Recognizing how it changes sign based when squared. This prevents misconceptions during arithmetic.
• Properly managing terms involving \( i \) in complex operations, allowing conversion between real and imaginary components.

Mastering the imaginary unit will unlock a deeper understanding of complex number operations as seen here.

Difference of Squares

The difference of squares is an algebraic identity especially useful when simplifying expressions with complex numbers. It states that \( a^2 - b^2 = (a + b)(a - b) \). This identity holds true for both real and complex numbers.
In the original problem, the denominator \( (6 - i)(6 + i) \) employs this identity. By recognizing \( (6 - i)(6 + i) \) as a difference of squares, it simplifies to \( 6^2 - i^2 \). What makes this special is the imaginary unit \( i \), where \( i^2 = -1 \), converting the expression to \( 36 + 1 = 37 \).
This approach achieves:

• Elimination of the imaginary part, reducing the expression to real numbers.
• Simplification of the denominator, which should always be a real number to ease the division.

Understanding the difference of squares serves as a powerful tool when dealing with complex expressions, providing clarity and reduction in complexity.