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Verified Answer :

kudzordzifrancis

ANSWER


The value of the expression is
\(- 1\)


EXPLANATION

Method 1: Rewrite as product of
\({i}^{2}\)


The expression given to us is,

\({i}^{0} \times {i}^{1} \times {i}^{2} \times {i}^{3} \times {i}^{4}\)


We use the fact that
\({i}^{2} = - 1\)
to simplify the above expression.



\({i}^{0} \times {i}^{1} \times {i}^{2} \times {i}^{3} \times {i}^{4} = {i}^{0} \times {i}^{1} \times {i}^{3} \times {i}^{2} \times {i}^{4}\)


This implies,


\({i}^{0} \times {i}^{1} \times {i}^{2} \times {i}^{3} \times {i}^{4} = {i}^{0} \times {i}^{2} \times {i}^{2} \times {i}^{2} \times {i}^{2} \times {i}^{2}\)


We substitute to obtain,

\({i}^{0} \times {i}^{1} \times {i}^{2} \times {i}^{3} \times {i}^{4} = 1\times - 1 \times - 1 \times - 1\times - 1 \times - 1\)


\({i}^{0} \times {i}^{1} \times {i}^{2} \times {i}^{3} \times {i}^{4} = 1\times 1 \times 1 \times - 1 = - 1\)


Method 2: Use indices to solve.



\({i}^{0} \times {i}^{1} \times {i}^{2} \times {i}^{3} \times {i}^{4} = {i}^{0 + 1 + 2 + 3 + 4}\)



This implies that,


\({i}^{0} \times {i}^{1} \times {i}^{2} \times {i}^{3} \times {i}^{4} = {i}^{10}\)




\({i}^{0} \times {i}^{1} \times {i}^{2} \times {i}^{3} \times {i}^{4} = ( {{i}^{2}} )^{5}\)


\({i}^{0} \times {i}^{1} \times {i}^{2} \times {i}^{3} \times {i}^{4} = ( - 1 )^{5} = - 1\)


BladeRunner212

- 1

Further explanation

This is a problem that is partly related to complex numbers, i.e., imaginary numbers. We will see how the power of i is an imaginary unit. Maybe we will see an interesting pattern.

\(\boxed{\boxed{ \ i = \sqrt{-1}\ }} \rightarrow \boxed{\boxed{ \ i^2 = -1 \ }}\)

Question:

The value of the expression \(\boxed{ \ i^0 \times i^1 \times i^2 \times i^3 \times i^4 \ }\)

The Process

Recall \(\boxed{ \ x^0 = 1 \ }\).

\(\boxed{ \ i^0 = 1 \ }\)

\(\boxed{ \ i^1 = \sqrt{-1} \ or \ i \ }\)

\(\boxed{ \ i^2 = (\sqrt{-1})^2 = -1 \ }\)

\(\boxed{ \ i^3 = i \times i^2 = i \times -1 = -i \ }\)

\(\boxed{ \ i^4 = i^2 \times i^2 = -1 \times -1 = 1\ }\)

Then \(\boxed{ \ i^0 \times i^1 \times i^2 \times i^3 \times i^4 = 1 \times i \times (-1) \times -i \times 1 \ }\)

\(\boxed{ \ i^0 \times i^1 \times i^2 \times i^3 \times i^4 = -1 \times -i^2 \ }\)

\(\boxed{ \ i^0 \times i^1 \times i^2 \times i^3 \times i^4 = i^2 \ }\)

The result is  \(\boxed{\boxed{ \ -1 \ }}\)

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Another method is to use the property of indices.

\(\boxed{ \ x^a \cdot x^b \rightleftharpoons x^{a+b} \ } \ and \ \boxed{ \ (x^a)^b) \rightleftharpoons x^{ab} \ }\)

\(\boxed{ \ i^0 \times i^1 \times i^2 \times i^3 \times i^4 = i^{0 + 1 + 2 + 3 + 4} \ }\)

\(\boxed{ \ i^0 \times i^1 \times i^2 \times i^3 \times i^4 = i^{10} \ }\)

\(\boxed{ \ i^0 \times i^1 \times i^2 \times i^3 \times i^4 = (i^2)^5 \ }\)

\(\boxed{ \ i^0 \times i^1 \times i^2 \times i^3 \times i^4 = {-1}^5 \ }\)

We get the same result, i.e., \(\boxed{\boxed{ \ -1 \ }}\)

- - - - - - -

Well, now pay attention to the pattern.

\(\boxed{ \ i^0 = 1 \ }\)

\(\boxed{ \ i^1 = i \ }\)

\(\boxed{ \ i^2 = -1 \ }\)

\(\boxed{ \ i^3 = -i \ }\)

\(\boxed{ \ i^4 = 1\ }\)

\(\boxed{ \ i^5 = i^2 \times i^3 = i \ }\)

\(\boxed{ \ i^6 = i^2 \times i^4 = -1 \ }\)

\(\boxed{ \ i^7 = i^2 \times i^5 = -i \ }\)

Pattern repeat every \(4^{th}\) power.

Learn more

  1. About complex numbers https://brainly.com/question/1658190
  2. The piecewise-defined functions brainly.com/question/9590016
  3. The composite function brainly.com/question/1691598

Keywords: what is the value of the expression, i⁰ × i¹ × i² × i³ × i⁴, imaginary number, unit, a complex number, the pattern, the property of indices

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