Problem 25 - Advanced Engineering Mathematics

Question

In Problems 1-26, write the given number in the form \(a+i b\). $$ \left(\frac{i}{3-i}\right)\left(\frac{1}{2+3 i}\right) $$

Step-by-Step Solution

Answer

The expression is \(\frac{9}{130} + i \frac{11}{130}\).

Step 1: Simplify the first fraction

To simplify \( \frac{i}{3-i} \), multiply the numerator and the denominator by the complex conjugate of the denominator, \( 3+i \). This yields:\[ \frac{i}{3-i} \times \frac{3+i}{3+i} = \frac{i(3+i)}{(3-i)(3+i)} \]This results in:\[ \frac{3i - 1}{10} \]Therefore, \( \frac{i}{3-i} = \frac{3i - 1}{10} \).

Step 2: Simplify the second fraction

Simplify \( \frac{1}{2+3i} \) by multiplying the numerator and the denominator by the complex conjugate of the denominator, \( 2-3i \):\[ \frac{1}{2+3i} \times \frac{2-3i}{2-3i} = \frac{2-3i}{(2+3i)(2-3i)} \]This results in:\[ \frac{2-3i}{13} \]Thus, \( \frac{1}{2+3i} = \frac{2-3i}{13} \).

Step 3: Multiply the fractions

Now, multiply the simplified forms of the two fractions:\[ \left( \frac{3i - 1}{10} \right) \times \left( \frac{2-3i}{13} \right) \]The numerator becomes \( (3i - 1)(2 - 3i) \), which after expansion is:\[ 6i - 9i^2 - 2 + 3i = 11i + 9 \]Therefore, the product is:\[ \frac{11i + 9}{130} \]

Step 4: Write in the form \(a+ib\)

The fraction \( \frac{11i + 9}{130} \) can be separated into real and imaginary parts:\[ a = \frac{9}{130} \quad \text{and} \quad b = \frac{11}{130} \]Thus, the expression in the form \( a + ib \) is:\[ \frac{9}{130} + i\frac{11}{130} \]

Key Concepts

Complex Conjugate

To simplify expressions involving complex numbers, we often use the concept of the complex conjugate. A complex conjugate of a complex number switches the sign of its imaginary part. If you have a complex number in the form \( a + bi \), its complex conjugate is \( a - bi \). This change helps to remove the imaginary part from the denominator when dealing with fractions.

Here's how it works:

• Find the complex conjugate of the denominator.
• Multiply both the numerator and the denominator by this conjugate.
• This results in a real number in the denominator after using the formula: \[ (a + bi)(a - bi) = a^2 + b^2 \]

Using the complex conjugate turns a messy division problem with complex numbers into one with a neat real denominator.

Real and Imaginary Parts

Every complex number is made up of two components: the real part and the imaginary part. Consider the number \( a + bi \), where \( a \) is the real part and \( bi \) is the imaginary part. Without understanding these parts, handling complex numbers can be tricky.

For example, in the expression \( \frac{11i + 9}{130} \), it's important to separate the real and imaginary parts for clarity.

• The real part here is \( \frac{9}{130} \).
• The imaginary part is \( i\frac{11}{130} \).

Distinguishing the real and imaginary parts simplifies the complex number handling and prepares the number for further operations or interpretations.

Multiplying Complex Numbers

Multiplying complex numbers might seem daunting, but it's straightforward with systematic steps. When you have complex numbers like \( (3i - 1)(2 - 3i) \), use the distributive property just like with binomials.

Here's the step-by-step breakdown:

• Multiply each part of the numbers, like \( 3i \times 2 \) and \( 3i \times -3i \).
• Remember that \( i^2 = -1 \), which simplifies any \( i^2 \) term to a real number.
• Combine like terms: \( 6i - 9(-1) - 2 + 3i \) becomes \( 11i + 9 \).

Once exploded into simpler terms, adding the like components gives a clear a + bi form. This method keeps operations manageable and resists algebraic mistakes.