Math Problem Statement
i^{27}^{}
Solution
To solve , we can use the properties of the imaginary unit .
First, recall the powers of :
i^1 & = i \\ i^2 & = -1 \\ i^3 & = -i \\ i^4 & = 1 \\ \end{aligned}$$ These powers repeat every four terms: $$i, -1, -i, 1$$. To find $$i^{27}$$, we can determine the remainder when 27 is divided by 4: $$27 \div 4 = 6 \text{ remainder } 3$$ So, $$i^{27}$$ is equivalent to $$i^3$$. From the list above: $$i^3 = -i$$ Therefore, $$i^{27} = \boxed{-i}$$. Would you like further details or have any questions? Here are 5 related questions you might find interesting: 1. What is $$i^{50}$$? 2. How does the pattern of powers of $$i$$ help in complex number calculations? 3. What is the significance of the imaginary unit $$i$$ in complex numbers? 4. How do you solve equations involving $$i$$? 5. What is the geometric interpretation of $$i$$ on the complex plane? **Tip:** When dealing with powers of $$i$$, find the remainder when the exponent is divided by 4 to simplify the expression.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Complex Numbers
Imaginary Unit
Exponentiation
Formulas
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Theorems
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Suitable Grade Level
Grades 9-12