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Chapter 1: Problem 49

Powers Evaluate the power, and write the result in the form \(a+b i\) $$(3 i)^{5}$$

Short Answer

Expert verified
The result is \(0 + 243i\).

Step by step solution

01

Express Power Using Exponent Rules

The given expression is \((3i)^5\). We can use the power rule \((a^b)^c = a^{bc}\) to simplify inside the brackets, noting that \(i^2 = -1\). First, express \((3i)^5\) as \((3^5) \cdot (i^5)\).
02

Calculate the Power of the Coefficient

Calculate \(3^5\). Start with \(3^5 = 3 \times 3 \times 3 \times 3 \times 3\). Calculate step-by-step: 1. \(3^2 = 9\)2. \(3^3 = 9 \times 3 = 27\) 3. \(3^4 = 27 \times 3 = 81\)4. \(3^5 = 81 \times 3 = 243\).
03

Calculate the Power of the Imaginary Unit

Calculate \(i^5\). Remember that the powers of \(i\) cycle every four terms:1. \(i^1 = i\)2. \(i^2 = -1\)3. \(i^3 = -i\)4. \(i^4 = 1\)and after that, it repeats. So, \(i^5\) is the same as \(i^1\), which is \(i\).
04

Combine Results to Find the Complex Number

Combine the results from steps 2 and 3. Multiply the result of \(3^5 = 243\) with the result of \(i^5 = i\):\((3i)^5 = 243 \times i = 243i\).The expression is already in the form \(a + bi\). Here, \(a = 0\) and \(b = 243\), so the expression is \(0 + 243i\).

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Key Concepts

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

Powers of i
When working with complex numbers, understanding the powers of the imaginary unit, denoted as \(i\), is essential. The imaginary unit \(i\) is defined as the square root of \(-1\), which is written as \(i^2 = -1\). This leads to an interesting pattern whenever you raise \(i\) to higher powers.

Here's the cyclical pattern of the powers of \(i\):
  • \(i^1 = i\): This is the value of \(i\) itself, not much transformation here.
  • \(i^2 = -1\): By definition, \(i\) denotes the square root of \(-1\), hence \(i^2 = -1\).
  • \(i^3 = -i\): Calculate \(i^3\) by multiplying \(i^2\) with \(i\). Hence, \((-1) \cdot i = -i\).
  • \(i^4 = 1\): Multiply \(i^3 = -i\) by \(i\) again, giving \(-i \cdot i = 1\).
These values repeat every four powers, creating a sustainable cycle: \(i, -1, -i, 1\), and back to \(i\). This pattern is crucial when working with higher powers, as you only need to find the remainder of the exponent divided by 4 to determine the power's effect.
Exponentiation
Exponentiation involves raising a number to the power of another. This concept is critical in mathematics and appears frequently in complex number calculations too. When you see an expression like \((a^b)^c\), using the rule \(a^{bc}\) simplifies it significantly.

Let's consider the number 3 raised to the power 5, written as \(3^5\). You calculate as follows:
  • Find \(3^2 = 9\).
  • Multiply again: \(9 \times 3 = 27\) yields \(3^3\).
  • Extend further: \(27 \times 3 = 81\), so you have \(3^4\).
  • Finally, \(81 \times 3 = 243\), bringing \(3^5\) to completion.
Exponentiation lets you express large numbers easily, enabling complex multiplication and simplifying mathematical models and predictions.
Complex Exponentiation
In complex exponentiation, we raise complex numbers to a power. The expression \((3i)^5\) is a perfect example. Here, 3 is the coefficient and \(i\) is the imaginary unit. For such expressions, each part鈥攔eal or imaginary鈥攕hould be examined separately.

Break it down:
  • Coefficient Calculation: First, compute the power of the coefficient, 3. We go from \(3^5\) to 243.
  • Imaginary Power: Then, calculate \(i^5\). Using the cycle (\(i, -1, -i, 1\)), we find that \(i^5\) is equivalent to \(i\).
Combine the results: Multiply the outcomes鈥擻(243\) and \(i\)鈥攖o get \(243i\). This final step results in the complex number in the form \(a + bi\), where \(a=0\), and \(b=243\). Complex exponentiation is a valuable tool in numerous mathematical areas, allowing advanced functionality in solving equations and analyzing periodic functions.

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Most popular questions from this chapter

The average height of adult males is 68.2 in., and \(95 \%\) of adult males have height \(h\) that satisfies the inequality $$ \left|\frac{h-68.2}{2.9}\right| \leq 2 $$ Solve the inequality to find the range of heights.

Show that the equation represents a circle, and find the center and radius of the circle. $$3 x^{2}+3 y^{2}+6 x-y=0$$

Write the number indicated in each statement in scientific notation. (a) A light-year, the distance that light travels in one year, is about \(5,900,000,000,000 \mathrm{mi}\) (b) The diameter of an electron is about \(0.0000000000004 \mathrm{cm}\) (c) A drop of water contains more than 33 billion billion molecules.

Area of a Region Find the area of the region that lies outside the circle \(x^{2}+y^{2}=4\) but inside the circle $$ x^{2}+y^{2}-4 y-12=0 $$

Radicals Simplify the expression, and eliminate any negative exponents(s). Assume that all letters denote positive numbers. (a) \(\sqrt{4 s t^{3}} \sqrt[6]{s^{3} t^{2}}\) (b) \(\frac{\sqrt[4]{x^{7}}}{\sqrt[4]{x^{3}}}\)
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