Problem 107 - Elementary and Intermediate Algebra

Question

Perform the operations. Write all answers in the form \(a+b i .\) $$ \frac{4}{5 i^{35}} $$

Step-by-Step Solution

Answer

The answer is \(0 + \frac{4}{5}i\).

Step 1: Simplify the Power of i

First, recall that the imaginary unit has the property that the powers of \(i\) cycle every four: \(i^1 = i,\) \(i^2 = -1,\) \(i^3 = -i,\) \(i^4 = 1,\) and then the cycle repeats. To simplify \(i^{35}\), find the remainder of \(35\div 4\). The quotient is 8 with a remainder of 3. Therefore, \(i^{35} = i^3 = -i\).

Step 2: Substitute Simplified Expression into the Fraction

Substitute \(i^{35}\) with \(-i\) from the simplification:\[\frac{4}{5i^{35}} = \frac{4}{5(-i)} = \frac{4}{-5i}.\]

Step 3: Rationalize the Denominator

To remove the imaginary unit from the denominator, multiply the numerator and the denominator by \(i\):\[\frac{4}{-5i} \times \frac{i}{i} = \frac{4i}{-5i^2}.\]Since \(i^2 = -1\), this becomes:\[\frac{4i}{5}.\] This is already in the form \(a + bi\), where \(a = 0\) and \(b = \frac{4}{5}\).

Key Concepts

Imaginary Unit

The imaginary unit, denoted as \(i\), is a fundamental concept in complex numbers. It is defined by the unique property that \(i^2 = -1\). This makes \(i\) quite different from ordinary numbers, as no real number squared results in a negative number. Thus, \(i\) helps in extending the set of real numbers to include solutions to equations like \(x^2 + 1 = 0\).
In the realm of complex numbers, any number can be expressed in the form \(a + bi\), where \(a\) and \(b\) are real numbers and \(bi\) represents the imaginary part. This form simplifies operations and problem-solving involving complex numbers, as seen in addition, subtraction, multiplication, and division.
Understanding and utilizing the imaginary unit effectively opens the door to solving more complex equations and understanding advanced mathematical concepts. It is crucial to remember that while \(i\)'s square gives \(-1\), other powers of \(i\) follow a cyclic pattern, which leads us to the next important concept.

Rationalizing Denominators

Rationalizing the denominator is a process in mathematics used to eliminate imaginary numbers from the denominator of a fraction. This technique is particularly useful because denominators without imaginary units help in further calculations and in maintaining a conventional expression of complex numbers.
To rationalize a denominator that contains \(i\), you typically multiply both the numerator and the denominator by the complex conjugate or directly by \(i\), as seeing in the fraction form. In the example we've been discussing, we have an expression \(\frac{1}{-5i}\). By multiplying the numerator and the denominator by \(i\), we convert it into a real number:

• \(\frac{4}{-5i} \times \frac{i}{i} = \frac{4i}{-5i^2}\)
• Using the property \(i^2 = -1\), this becomes \(\frac{4i}{5}\), which is simpler.

This method is particularly efficient for maintaining a standard form, \(a + bi\), where it becomes easier to identify both the real and the imaginary components. This simplifies both calculation and comprehension of complex numbers.

Powers of i

The powers of the imaginary unit \(i\) follow a cyclic pattern every four exponents, which is pivotal in simplifying expressions involving complex numbers. To understand this, let's review the cycle:

• \(i^1 = i\)
• \(i^2 = -1\)
• \(i^3 = -i\)
• \(i^4 = 1\)

After these four, the pattern repeats: \(i^5 = i\), \(i^6 = -1\), and so forth. This cyclicity simplifies the computation of powers of \(i\), as in our example, where we calculated \(i^{35}\).
By finding the remainder of \(35\div4\), we determine that \(i^{35}\) is equivalent to \(i^3 = -i\). An understanding of this repeating cycle is essential in breaking down complex numbers into manageable parts, aiding in further mathematical operations.
Recognizing this cyclical nature not only simplifies calculations but also deepens the comprehension of complex numbers and their properties. It enables mathematicians and students alike to perform operations on complex numbers efficiently and accurately.