Problem 46 - Precalculus Mathematics for Calculus

Question

Evaluate the expression and write the result in the form \(a+b i\) $$\frac{(1+2 i)(3-i)}{2+i}$$

Step-by-Step Solution

Answer

The expression evaluates to \(3 + i\).

Step 1: Expand the Numerator

First, expand the product in the numerator. Multiply \((1 + 2i)\) by \((3 - i)\).\((1 + 2i)(3 - i) = 1 \cdot 3 + 1 \cdot (-i) + 2i \cdot 3 + 2i \cdot (-i) = 3 - i + 6i - 2i^2\).Since \(i^2 = -1\), we have:\(3 - i + 6i + 2 = 5 + 5i\).

Step 2: Multiply by the Conjugate of the Denominator

Next, deal with the denominator by multiplying both the numerator and the denominator by the conjugate of the denominator. The conjugate of \(2 + i\) is \(2 - i\).Multiply the expression by \(\frac{2-i}{2-i}\):\(\frac{(5+5i)(2-i)}{(2+i)(2-i)}\).

Step 3: Simplify the Denominator

Simplify the denominator by multiplying:\((2+i)(2-i) = 2^2 - (i^2) = 4 - (-1) = 5\).

Step 4: Expand the Adjusted Numerator

Expand the numerator by multiplying:\((5+5i)(2-i) = 5 \cdot 2 + 5 \cdot (-i) + 5i \cdot 2 + 5i \cdot (-i) = 10 - 5i + 10i - 5i^2\).Simplify using \(i^2 = -1\):\(10 - 5i + 10i + 5 = 15 + 5i\).

Step 5: Divide Terms by the Real Number

Finally, divide each term of the numerator by the denominator which is a real number:\(\frac{15+5i}{5} = \frac{15}{5} + \frac{5i}{5} = 3 + i\).This is the expression in the form \(a + b i\).

Key Concepts

Imaginary Unit

The imaginary unit, denoted by \(i\), is a fundamental concept in complex numbers. It's defined as the square root of -1, which means \(i^2 = -1\). This allows us to work with expressions involving roots of negative numbers, which aren’t solvable in the realm of real numbers.

When you combine real numbers with imaginary numbers like \(i\), you form complex numbers. These numbers take the format \(a + bi\), where \(a\) is the real part and \(b\) is the imaginary part.

• For example, in the expression \((1+2i)\), the coefficient \(2\) is multiplied by \(i\) to express the imaginary part.
• This component allows for calculations that involve both real and imaginary quantities, expanding the possibilities beyond the real number line.

Understanding \(i\) is key to manipulating and simplifying complex expressions effectively.

Conjugate

The conjugate of a complex number is obtained by reversing the sign of its imaginary component. Given a complex number \(a + bi\), its conjugate is \(a - bi\). This is crucial because it helps simplify division involving complex numbers.

In our exercise, the denominator \(2+i\) has its conjugate \(2-i\). By multiplying both the numerator and the denominator by \(2-i\), we employ this concept to eliminate the imaginary part in the denominator.

• Multiplying by the conjugate ensures that the denominator becomes a real number, which simplifies further calculations.
• For example, \((2+i)(2-i)=4-(-1)=5\). This clear result is possible because the conjugate creates a difference of squares.

This technique is a key step in expressing complex fractions in their standard form \(a + bi\).

Simplification

Simplification in the context of complex numbers involves reducing the expression to its most straightforward form. This process includes collecting like terms and using properties of \(i\).

During simplification, terms involving \(i^2\) are converted using \(i^2 = -1\). For instance, in the numerator calculation, \(-2i^2\) becomes \(+2\) because \(-2(-1)=+2\).

• After expanding and combining like terms, expressions are organized into real and imaginary components.
• Simplification makes complex calculations more manageable, forming the expression \(5 + 5i\) after the initial expansion.

Properly applying these simplifications is essential for reaching the final form efficiently.

Numerator and Denominator

Understanding the roles of the numerator and denominator in a complex fraction is vital, as they guide the simplification process. When you have a complex fraction, you work on separating the imaginary and real parts efficiently.

In this exercise, the initial step involves expanding the numerator by multiplying the two binomials: \((1+2i)\) and \((3-i)\). This expansion results in a new expression that combines the real and imaginary parts.

Similarly, the denominator is initially tackled by multiplying the complex number by its conjugate. This clears the imaginary unit from the denominator, turning it into a real number.

• The numerator becomes \(15 + 5i\) after expansion and simplification.
• The denominator becomes \(5\) after multiplying by the conjugate.

Finally, dividing the simplified numerator by the real number denominator leads to the solved form. This showcases how both parts play a distinct role in arriving at the final expression in \(a + bi\) form.