Falling_In_Katt
  • Falling_In_Katt
I^15 =
Mathematics
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Falling_In_Katt
  • Falling_In_Katt
I^15 =
Mathematics
katieb
  • katieb
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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welshfella
  • welshfella
i^2 = -1 , i^3 = ii and i^4 = 1
welshfella
  • welshfella
* i^3 = -i
welshfella
  • welshfella
i^12 = (i^4)^3 = 1 i^15 = i^4 * i^3

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welshfella
  • welshfella
* last line is i^15 = i^12 * i^3
Falling_In_Katt
  • Falling_In_Katt
I'm so confused.
welshfella
  • welshfella
i^2 = -1 thats by definition right?
Falling_In_Katt
  • Falling_In_Katt
wait...would it be -i
marihelenh
  • marihelenh
1 raised to any power is always going to be 1. You could right it out as 1*1*1*1*1*1*1*1*1*1*1*1*1*1*1=1 if you wanted, but it will always be 1.
welshfella
  • welshfella
yes -i is correct
Falling_In_Katt
  • Falling_In_Katt
Oh thank you!
Michele_Laino
  • Michele_Laino
we have: \[{i^2} = - 1 \Rightarrow {i^{14}} = {\left( {{i^2}} \right)^7} = {\left( { - 1} \right)^7} = - 1 \Rightarrow {i^{15}} = {i^{14}}i = ...?\]
welshfella
  • welshfella
i assumed this was an i - in the question - but it looks like 1 lol!

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