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Chapter 7: Problem 25

Write each expression in the form \(a+b i,\) where a and b are real numbers. \(i^{8001}\)

Short Answer

Expert verified
The expression \(i^{8001}\) in the form \(a+bi\) is \(0+1i\).

Step by step solution

01

Determine the exponent modulo 4

Since the powers of i repeat in a cycle of every 4, we can get the remainder when 8001 is divided by 4. So, we have \(8001\text{ mod }4 = 1\), because \(8001 = 4 \cdot 2000 + 1\).
02

Find the value of i raised to the exponent

Since the remainder is 1, this means the value of \(i^{8001}\) is equal to the value of \(i^1\), which is simply \(i\). Therefore, \(i^{8001} = i\).
03

Express the result in the form a+bi

Now that we know \(i^{8001} = i\), we can express this in the form \(a+bi\), where \(a\) and \(b\) are real numbers. In this case, \(a = 0\) and \(b = 1\), so the expression becomes: \[i^{8001} = 0 + 1i.\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Imaginary Unit
The concept of the imaginary unit is fundamental in understanding complex numbers. The imaginary unit is denoted by the symbol \(i\), and its primary property is that \(i^2 = -1\). This property is what differentiates imaginary numbers from real numbers. When you multiply \(i\) by itself, you essentially step into a new dimension of numbers that aren't directly compatible with our normal number line used for reals, which extends to the inclusion of complex numbers.
  • \(i = \sqrt{-1}\)
  • \(i^2 = -1\)
  • \(i^3 = i \cdot i^2 = -i\)
  • \(i^4 = i^2 \cdot i^2 = 1\)
Each time \(i\) is raised to a power, its result cycles back through these values. This cyclical nature is a key aspect to solving expressions where an imaginary unit is raised to a high power.
Powers of i
Understanding the powers of \(i\) involves recognizing the repeating pattern that arises from successive powers of the imaginary unit. As seen, the powers of \(i\) repeat every four cycles:
  • \(i^1 = i\)
  • \(i^2 = -1\)
  • \(i^3 = -i\)
  • \(i^4 = 1\)
After \(i^4\), the cycle starts over, as \(i^5 = i^1\), and so on. This cycle allows us to simplify expressions like \(i^{8001}\) efficiently. By performing division of the exponent by 4 and finding the remainder, as done with 8001 mod 4 = 1, we identify which part of the cycle these high powers fall into. Thus, \(i^{8001} = i^1 = i\).
Modular Arithmetic
Modular arithmetic is a system of arithmetic for integers, where numbers 'wrap around' upon reaching a certain value鈥攖he modulus. It's akin to the arithmetic of a clock, where after 12, it wraps back to 1. In the context of powers of \(i\), modular arithmetic helps simplify the calculations greatly. Recognizing that powers of \(i\) repeat every four terms means that we can use a modulus of 4.
  • This approach reduces complexity, as seen in the calculation \(8001 \mod 4 = 1\).
  • By knowing \(8001 = 4 \times 2000 +1\), the focus is narrowed to \(i^1\), thus simplifying \(i^{8001}\) to just \(i\).
  • Without modular arithmetic, calculating such high powers would be impractical.
Using modular arithmetic not only streamlines the process but also eliminates the need for exhaustive computation.

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Most popular questions from this chapter

Show that if \(\mathbf{u}, \mathbf{v}\) and \(\mathbf{w}\) are vectors, then $$ \mathbf{u} \cdot(\mathbf{v}+\mathbf{w})=\mathbf{u} \cdot \mathbf{v}+\mathbf{u} \cdot \mathbf{w} $$ .

Show that \(\overline{w-z}=\bar{w}-\bar{z}\) for all complex numbers \(w\) and \(z\)

Describe the subset of the complex plane consisting of the complex numbers \(z\) such that the real part of \(z^{3}\) is a positive number.

Suppose $$ f(x)=a x^{2}+b x+c $$ where \(a \neq 0\) and \(b^{2}<4 a c .\) Verify by direct substitution into the formula above that $$ f\left(\frac{-b+\sqrt{4 a c-b^{2}} i}{2 a}\right)=0 $$ and $$ f\left(\frac{-b-\sqrt{4 a c-b^{2}} i}{2 a}\right)=0 $$

Explain why the six distinct complex numbers that are sixth roots of 1 are the vertices of a regular hexagon inscribed in the unit circle.
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