Problem 19 - Advanced Engineering Mathematics

Question

Write the given number in the form \(a+i b\). $$ \frac{(5-4 i)-(3+7 i)}{(4+2 i)+(2-3 i)} $$

Step-by-Step Solution

Answer

The expression is \(\frac{23}{37} - \frac{64}{37}i\).

Step 1: Simplify the Numerator

First, we simplify the expression in the numerator \((5-4i) - (3 + 7i)\). Begin by distributing the negative sign in front of the second term: \(5-4i - 3 - 7i\). This simplifies to \((5-3) + (-4i - 7i)\), resulting in \(2 - 11i\).

Step 2: Simplify the Denominator

Next, we simplify the expression in the denominator \((4+2i) + (2-3i)\). Combine like terms: \((4+2) + (2i-3i)\), resulting in \(6 - i\).

Step 3: Write as a Fraction

Now write the simplified expressions as a fraction: \(\frac{2 - 11i}{6 - i}\).

Step 4: Multiply by the Conjugate

To eliminate the imaginary part from the denominator, multiply both the numerator and the denominator by the conjugate of the denominator, \(6 + i\). The expression becomes: \(\frac{(2 - 11i)(6 + i)}{(6 - i)(6 + i)}\).

Step 5: Multiply Out the Numerator

Expand \((2 - 11i)(6 + i)\). Apply distributive property: \(2 \times 6 + 2 \times i - 11i \times 6 - 11i \times i\). This results in \(12 + 2i - 66i - 11i^2\). Since \(i^2 = -1\), replace \(-11i^2\) with \(11\): gives \(12 + 2i - 66i + 11\), which simplifies to \(23 - 64i\).

Step 6: Multiply Out the Denominator

Expand \((6 - i)(6 + i)\). Using the difference of squares: \(6^2 - (i)^2\) results in \(36 - (-1) = 36 + 1 = 37\).

Step 7: Simplify the Expression

Now our expression is \(\frac{23 - 64i}{37}\). Divide each term in the numerator by 37: \(\frac{23}{37} - \frac{64i}{37}\). Thus, the expression in the form of \(a + ib\) is \(\frac{23}{37} - \frac{64}{37}i\).

Key Concepts

Imaginary Unit

In mathematics, the imaginary unit is denoted by the symbol \(i\). It represents the square root of -1. This concept allows us to extend our number system so that we can solve equations that have no real solutions like \(x^2 + 1 = 0\). When dealing with the imaginary unit:

• \(i^2\) equals -1.
• \(i^3\) equals -i.
• \(i^4\) equals 1.

These properties allow us to simplify expressions involving powers of \(i\). In complex numbers, \(i\) enables us to incorporate imaginary parts into real numbers to form expressions like \(a+bi\). Complex numbers are foundational for many aspects of science and engineering. They allow for the description of phenomena that cannot rely solely on real numbers.

Conjugate

In the realm of complex numbers, the conjugate is an essential concept. For any complex number \(a + bi\), its conjugate is \(a - bi\). The idea is simple but powerful:

• The conjugate changes the sign of the imaginary part.
• It is useful for removing imaginary parts from denominators.
• The product of a complex number and its conjugate is always a real number, acting as a difference of squares: \((a+bi)(a-bi) = a^2 - (bi)^2 = a^2 + b^2\).

When you multiply a complex number by its conjugate, you essentially rationalize any imaginary number, making divisions involving complex numbers feasible and straightforward.

Rationalizing Denominator

Rationalizing the denominator is a method used to eliminate the imaginary unit from the denominator of a fraction. This enables expressions to be represented in a standard format, usually making them more interpretable. When it comes to complex numbers, we achieve this by multiplying both the numerator and the denominator by the conjugate of the denominator. Let's illustrate:

• Given a fraction, for instance, \(\frac{2 - 11i}{6 - i}\), identifying the denominator's conjugate gives \(6 + i\).
• Multiply the entire expression by \(\frac{6 + i}{6 + i}\). This operation results in a denominator that is a real number.

This process is essential because it simplifies complex number expressions into the standard form \(a + bi\), which is more straightforward to understand and work with.

Complex Number Division

Dividing complex numbers, like \(\frac{2 - 11i}{6 - i}\), involves a few crucial steps to ensure a neat answer in the form \(a + bi\). After recognizing that direct division is not possible due to the imaginary part in the denominator, we use the conjugate to simplify:

• First, simplify both the numerator and denominator individually if needed.
• Multiply both by the conjugate of the denominator, transforming the division into a multiplication that cancels out the imaginary units in the denominator.
• After multiplying, simplify both the new numerator and denominator, transforming them into a real number and an imaginary number.

Ultimately, complex number division enables expressions with imaginary numbers to be simplified and rewritten neatly as real and imaginary components. This method enhances calculation accuracy and expression clarity.