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i^21+i^32-i^67 =
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@hartnn
i=?
i=underroot(-1)
i^2=?
-1
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i^3=?
i^3=?
i^4=?
i^3=-i ,i^4=1
i^5=? so u see the pattern ?
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I understand the pattern but when i get higher powers i get confused :(
\(\huge i=\sqrt{-1} \\ \huge i^2=-1 \\ \huge i^3=-i \\ \huge i^4=1 \\~ \\~ \huge i^5=\sqrt{-1} \\ \huge i^6=-1 \\ \huge i^7=-i \\ \huge i^8=1\\~\\~ \huge i^9=\sqrt{-1} \\ \huge i^{10}=-1 \\ \huge i^{11}=-i \\ \huge i^{12}=1\)
so i raised to any 4th power =1
write i^21 as i*i^20 what is i^20 = ?
1 ?
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so i^21 = ?
i
yes, what about i^32 = ?
1
try i^67
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i^3
thats correct, and i^3= ?
-underroot(-1)
thats -i and final answer = ?
1
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yup. clear with all steps ?
yeah thanksalot :)
wait... i+1-(-i)=?
No it is a plus not a minus sorry :)
I wrote the question wrong
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then its 1
Yeah :)
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