simplify \[i ^{67}\]
ok just like before. divide 67 by 4 and take the remainder
4|67 R... youre getting quicker ;)
lets call the remainder r then \[i^{67}=i^r\]
i got 16.74
i meen i got 16.75
hold the phone
=i^3=-i
its not 16.75 .. its either 1, -1, i or -i
i mean divide, get whole number and take the remainder
like if i divide 72 by 5 i get a remainder of 2
i did not mean divide and get a decimal
the reason we divide by 4 is that there is only 4 answers it can be; they just circle around on each other like the hands of a clock
if i divide 111 by 10 i get a remainder of 1 and if you divide 67 by 3 the remainder is...
so the answere is i or 1?
its got to be 1 then
slow
67/4 = 16 R3. i^3 = i^0 = 1 i^1 = i i^2 = -1 i^4 = -i
\[i^1=i\] \[i^2=-1\] \[i^3=-i\] \[i^4=1\]
.... i have cursed fingers lol
it is from this set \[\{1, i, -1, -i\}\] depending on the remainder when you divide your exponent by 4
if the remainder is 1, you get \[i^1=i\] if it is 2 you get \[i^2=-1\] if 3 you get \[i^3=-i\] and if 4 divides your exponent evenly you get \[i^0=1\]
ok got it!
good. clear what remainder means yes?
\[i^{67}=i^{4\cdot16+3}=i^{4\cdot16}i^3=(i^4)^{16}i^3=1^{16}i^3=i^3\]
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