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hartnn (hartnn):
you know whats i ?
OpenStudy (anonymous):
variable?
hartnn (hartnn):
you are in imaginary numbers topic , right ?
so, \(\large i=\sqrt{-1}\)
heard of it ?
OpenStudy (anonymous):
Yeah i i heard of it
hartnn (hartnn):
so, whats i^2 =... ?
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OpenStudy (anonymous):
1 ?
hartnn (hartnn):
no!
\(i=\sqrt {-1}\)
squaring both sides,
\(i^2=-1\)
got this ?
OpenStudy (anonymous):
ok
hartnn (hartnn):
to get i^3, just multiply i on both sides
i^3=-i
ok?
OpenStudy (anonymous):
ohh ok soo like i^3(-1) or somthing like that?
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hartnn (hartnn):
what ?
i^3 = -i
OpenStudy (anonymous):
sorry its really new for me kinda lost
hartnn (hartnn):
i = sqrt (-1)
so i^2 = -1
i^3 = i*i^2 = i*-1 =-i
which part did u not get ?
OpenStudy (anonymous):
if i wanted to find i^4= i^3*i=i*-1
or i^2*i^2=i^2*-1?
hartnn (hartnn):
both way works but use that i^3=-i
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hartnn (hartnn):
so, i^4 = i^2*i^2 = -1 *-1 = 1
got this ?
OpenStudy (anonymous):
Ohh ok i^5= i^2*i*3= -1*-i=i?
hartnn (hartnn):
corrct! :)
OpenStudy (anonymous):
What if i were to do a number like i^36?
hartnn (hartnn):
we note the pattern,
like i^4 =1, we will get i^8 also =1
i^12 = 1 , i^16 =1 and so on...
do u notice a pattern ?
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OpenStudy (anonymous):
Yes soo like i^3=-i and i^9= -1 and every like multypul of 3 will be -i
hartnn (hartnn):
nopes, that worked for multiples of 4
i^3 = i^7 = i^11 =i^15 =... = -i
means if the exponents are \(\huge 4 \) apart, the value is same
hartnn (hartnn):
similarly,
i^2=i^6=i^10=...= -1
OpenStudy (anonymous):
ok i see was about to say same thing lol
hartnn (hartnn):
cool, so u now understand ?
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OpenStudy (anonymous):
yeah im getting it more and more soo i^36= 1
hartnn (hartnn):
correct! :)
can i ask you one ?
i^22 = ... ?
OpenStudy (anonymous):
ok -1
hartnn (hartnn):
good! :)
hartnn (hartnn):
want few more ?
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OpenStudy (anonymous):
i^2=-1 i^6=-1......22^i=-1 sure yeah
ganeshie8 (ganeshie8):
wat about i^4000000000000001
:)
hartnn (hartnn):
lol, i was trying to be bit lineant and ask him i^1001
hartnn (hartnn):
i^4000000000000001 =... ?
i^1001 =.... ?
its easy once you get that i^4 =1
hartnn (hartnn):
****
\(\large i^{4n}=1\)
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OpenStudy (anonymous):
ლ(゚д゚ლ)
hartnn (hartnn):
for any integer n
ganeshie8 (ganeshie8):
that looks like one of i's lol :P
OpenStudy (anonymous):
just so ik i^5 =i^3*i^2=i?
ganeshie8 (ganeshie8):
first kill the i^4 s
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hartnn (hartnn):
i^5 = i^4 * i = i
better ?
OpenStudy (anonymous):
ohh haha yes
OpenStudy (anonymous):
i=-1 if it didnt have a right
OpenStudy (anonymous):
or just i
hartnn (hartnn):
what ?
i = sqrt -1
i^2=-1
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OpenStudy (anonymous):
ohh yeah yeah i frgot
OpenStudy (anonymous):
what questions were u going to throw at me ?
hartnn (hartnn):
i^4000000000000001 =... ?
i^1001 =.... ?
hartnn (hartnn):
lets take i^101 first,
how would u go on solving this ?
you know i^4 =1 , so is i^8, so is i^12 and so on...actually when powers are multiples of 4, we get answer =1
so why not separate a power of 4 from 101 ?
100 divisible by 4 right ? so i^100 = 1
so, i^101 = i^100 *i = 1*i=i
got this ?
OpenStudy (anonymous):
ok i see how you got i^100=4 i^101 =?
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OpenStudy (anonymous):
ohh waut nvm i^101 =1
hartnn (hartnn):
no....i^100 = 1
because 100 is multiple of 4
OpenStudy (anonymous):
ohh yeah my bad i messed up typeing that how did u get i^101?
hartnn (hartnn):
i^101 is just i^100 times i
OpenStudy (anonymous):
ok ok i^4000000000000001 =... i^400000000000000 *1^1 would i do that
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hartnn (hartnn):
i^400000000000000 *i^1
yes!
i^400000000000000 *i^1 = 1*i^1 = i
hartnn (hartnn):
wasn't that easy ?
OpenStudy (anonymous):
yeah kinda
hartnn (hartnn):
same way i^1001 = .... ?
i promise, this is last :P
OpenStudy (anonymous):
ok i^1000=1*i=i
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hartnn (hartnn):
i^1001 = i^1000*i^1 = 1*i =i
correct! good work :)
hartnn (hartnn):
i hope you are more confident on this topic now :)