Question

What is the value of the expression i 0 × i 1 × i 2 × i 3 × i 4?

a.1
b.-1
c.i
d.-i

Answer 1

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,


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




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


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

Answer 2

- 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  $\boxed{\boxed{ \ -1 \ }}$

- - - - - - -

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