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Simplify i^16 A)-1 B)1 C)i D)-i
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Hint: \[i^2 = -1\]
X it by 16?
No, what if I write it as \[i^{16} = i^2 \times i^2 \times i^2....\] 8 times
\[(i^2)^8 =~?\]
\[(i^4)^4\]
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1?
Yup!
\[(i^2)^{32/4}\]
Lol ok im done.
lol
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jhannybean don't know what you meant on that last comment you put lol
Oh, i said im done trying to find derivations of the same thing : \(i^{16}\) lol.
I was talking about where you got 32/4
any exponent of i divisible by 4=1 ie i^16, i^32, i^400 all = 1
I would had never known that if you did not just tell me so thank you:)
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:)
32/4 = 8
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