Problem 18 - Precalculus: Functions and Graphs

Question

Write the expression in the form \(a+b i,\) where \(a\) and \(b\) are real numbers. $$\begin{array}{lll}\text { (a) } i^{\text {66}} & \text { (b) } i^{-55}\end{array}$$

Step-by-Step Solution

Answer

(a) -1 + 0i, (b) 0 + 1i

Step 1: Understanding Powers of i

The imaginary unit, denoted as i, has a defined cyclic pattern, particularly when raised to integer powers. Specifically, it behaves cyclically every four exponents as follows:- \(i^1 = i\)- \(i^2 = -1\)- \(i^3 = -i\)- \(i^4 = 1\).This cycle repeats for all higher powers of i.

Step 2: Calculating i^66

To find \(i^{66}\), determine where 66 falls in the cycle of four powers. Compute the remainder of the division of 66 by 4: \(66 \mod 4 = 2\). This remainder tells us that \(i^{66} = i^2 = -1\).Thus, the expression for \(i^{66}\) in the form \(a + bi\) is \(-1 + 0i\), which means \(a = -1\) and \(b = 0\).

Step 3: Calculating i^-55

For negative powers of i, utilize the same cycle of four, knowing that \(i^{-n} = (i^n)^{-1}\). First, find out what \(i^{55}\) is within the cycle:Compute \(55 \mod 4 = 3\), indicating \(i^{55} = i^3 = -i\). Thus, \(i^{-55} = (i^{55})^{-1} = (-i)^{-1}\).The inverse of \(-i\) is \(\frac{1}{-i}\). To rationalize, multiply by the conjugate:\[-\frac{i}{i^2} = \frac{-i}{-1} = i\].Thus, \(i^{-55} = i \), which in the form \(a + bi\) is \(0 + 1i\), meaning \(a = 0\) and \(b = 1\).

Key Concepts

Imaginary Unit

The imaginary unit, frequently denoted as \(i\), is a fundamental component of complex numbers. It is defined by the property that \(i^2 = -1\). This unique property differentiates it from real numbers and is the key to working with complex numbers.
In mathematical terms, the introduction of \(i\) allows us to solve equations like \(x^2 + 1 = 0\), which wouldn't have real number solutions. This gives complex numbers a richer space in which they can exist.
In practical applications, complex numbers are used in fields like engineering, physics, and applied mathematics to study phenomena such as electrical currents, signal processing, and quantum mechanics. They add extra dimensions of analysis and help us solve real-world problems.

Powers of i

The powers of the imaginary unit \(i\) follow a predictable cycle that repeats every four exponents. This cycle is crucial for simplifying expressions containing higher powers of \(i\):

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

This means that for any integer \(n\), the value of \(i^n\) can be found by determining the remainder when \(n\) is divided by 4. For instance, to find \(i^{66}\), compute \(66 \mod 4\) to get 2. Thus, \(i^{66} = i^2 = -1\).
Understanding this cycle is vital because it simplifies calculations involving complex numbers, removing the need to manually multiply \(i\) repeatedly.

Cyclic Pattern of i

The cyclic pattern of \(i\) is an efficient way to work with powers of \(i\) without intensive computation. This pattern is inherent due to how \(i\) interacts with multiplication. After each four multiples of \(i\), the cycle starts over.
If the exponent is negative, the cycle concept still applies. For example, to solve for \(i^{-55}\), find the equivalent positive power \(i^{55}\). This is done by computing \(55 \mod 4\) to get 3, hence \(i^{55} = i^3 = -i\).
Even for negative exponents, we convert them to positive by finding the multiplicative inverse. Therefore \(i^{-55} = (i^{55})^{-1} = (-i)^{-1} = i\). This cyclic understanding reinforces the simplicity of dealing with complex number powers, streamlining problem-solving processes.