Problem 36 - Algebra and Trigonometry

Question

Evaluate the expression and write the result in the form a bi. $$ \frac{1}{1+i} $$

Step-by-Step Solution

Answer

The result is \(\frac{1}{2} - \frac{i}{2}\).

Step 1: Identify the Conjugate

The given expression to evaluate is \( \frac{1}{1+i} \). To simplify this, we need to remove the imaginary unit \(i\) from the denominator. This can be achieved by multiplying both the numerator and denominator by the conjugate of \(1+i\), which is \(1-i\).

Step 2: Multiply by the Conjugate

Multiply both the numerator and the denominator by \(1-i\):\[\frac{1}{1+i} \times \frac{1-i}{1-i} = \frac{1 \times (1-i)}{(1+i)(1-i)}\]

Step 3: Compute the Denominator

Calculate the denominator \((1+i)(1-i)\) by using the difference of squares formula:\[(1+i)(1-i) = 1^2 - i^2 = 1 - (-1) = 1 + 1 = 2\]

Step 4: Simplify the Numerator

The numerator becomes:\[1 \times (1-i) = 1-i\]

Step 5: Write the Simplified Expression

Combine the results of the numerator and denominator:\[\frac{1-i}{2} = \frac{1}{2} - \frac{i}{2}\]Therefore, the expression in the form \(a + bi\) is \(\frac{1}{2} - \frac{i}{2}\).

Key Concepts

Conjugate

In the realm of complex numbers, the conjugate is a crucial concept that aids in simplifying expressions. The conjugate of a complex number involves changing the sign of its imaginary part, while the real part remains the same. For example, if we have a complex number of the form \( a + bi \), its conjugate would be \( a - bi \). Simply put, the conjugate flips the sign of the imaginary component.
In our original exercise, we looked at the number \( 1+i \). The conjugate here, therefore, is \( 1-i \). By multiplying a complex number by its conjugate, we can effectively eliminate the imaginary unit from the denominator, simplifying complex fractions. This process is fundamental in presenting results in the format \( a + bi \), where \( a \) and \( b \) are real numbers.
Understanding conjugates is vital not only for solving equations, but also for working with roots in complex numbers, and performing division between complex numbers. Using conjugates thus becomes a powerful tool in many calculations involving complex numbers.

Imaginary Unit

The imaginary unit, denoted as \( i \), is defined by the property \( i^2 = -1 \). This definition is what sets the stage for the creation of complex numbers. Complex numbers are generally expressed as \( a + bi \), where \( a \) and \( b \) are real numbers. The term \( bi \) represents the imaginary component of the complex number.
Working with the imaginary unit is a bit different than working with real numbers. For instance:

• \( i \) is the basic imaginary unit.
• \( i^2 = -1 \), highlighting that multiplying \( i \) by itself results in a real number.
• Higher powers of \( i \) cycle through a pattern due to \( i^2 = -1 \): \( i^3 = -i \), \( i^4 = 1 \), and then repeating.

In our exercise, eliminating \( i \) from the denominator is a key step in simplifying the expression. This is made possible by multiplying the numerator and denominator of a fraction by the complex conjugate. In this way, understanding the imaginary unit and its properties is essential in complex number arithmetic.

Difference of Squares Formula

The difference of squares is a notable algebraic identity, and it is especially useful when working with complex numbers. This formula states that for any two numbers \( a \) and \( b \), the expression \( a^2 - b^2 \) can be rewritten as \((a+b)(a-b)\).
In the context of complex numbers, this formula helps in simplifying expressions that involve conjugates. When you multiply a complex number by its conjugate, you apply the difference of squares to get \[ (a+bi)(a-bi) = a^2 - (bi)^2 = a^2 - b^2i^2 \] Since \( i^2 = -1 \), the expression becomes \[ a^2 + b^2 \]
In our example, when we multiply \( (1+i) \) by its conjugate \( (1-i) \), the result is \( 1^2 - (i)^2 = 1 - (-1) \) which simplifies to \( 1+1 = 2 \). This operation effectively removes the imaginary unit from the denominator, thereby simplifying the fraction.
This algebraic identity not only simplifies fractions but also plays a significant role in calculus, polynomial division, and other areas of advanced mathematics.