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If [tex]i = \sqrt{-1}[/tex], then [tex]\frac{i + i^2 + i^3}{i^3 + i^4 + i^5} =[/tex] ?

A. -3
B. -1
C. [tex]\frac{1}{2}[/tex]
D. 1
E. 3


Sagot :

To solve the expression [tex]\(\frac{i + i^2 + i^3}{i^3 + i^4 + i^5}\)[/tex] where [tex]\(i = \sqrt{-1}\)[/tex], we need to evaluate the powers of [tex]\(i\)[/tex].

1. First, recall that the imaginary unit [tex]\(i\)[/tex] satisfies:
- [tex]\(i^2 = -1\)[/tex]
- [tex]\(i^3 = -i\)[/tex]
- [tex]\(i^4 = 1\)[/tex]
- [tex]\(i^5 = i\)[/tex] (because [tex]\(i^5 = i \cdot i^4 = i \cdot 1 = i\)[/tex])

2. Substitute these values into the numerator and denominator of the given expression.

3. Evaluate the numerator:
[tex]\[ i + i^2 + i^3 = i + (-1) + (-i) = i - 1 - i = -1 \][/tex]

4. Evaluate the denominator:
[tex]\[ i^3 + i^4 + i^5 = (-i) + 1 + i = -i + 1 + i = 1 \][/tex]

5. Now, we calculate the fraction:
[tex]\[ \frac{i + i^2 + i^3}{i^3 + i^4 + i^5} = \frac{-1}{1} = -1 \][/tex]

Thus, the value of the given expression is [tex]\(\boxed{-1}\)[/tex].
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