simplify i^127?

Answer 1

You can find a hundrend ways to obtain this and all are ok,but here's a simple general rule for evaluating those things

i^x=?

Make the division of THE LAST TWO DIGITS of x by 4...
(If x has one digit,then divide x by 4).
The remainder will be 0,1,2 or 3 (let's call it 'r').
If r=0 then i^x= 1
If r=1 then i^x= i
If r=2 then i^x=-1
If r=3 then i^x= -i

I think it's simpler that way.
Very simpler actually:
take for example i^124355543523255362.
In twenty seconds you can see that the remainder from the division of 62 by 4 is 2,therefore
i^124355543523255362=-1.
If you try another way,it will take you much more!


In this case,the remainder of 27/4 is 3
(27=4*6+3), so i^127=-i

P.S. That thing that
"rumplestiltskin12357" is saying (about even and odd powers) is completely wrong.

P.S.2.The idea behind the proof is that i^(4q+r)=...=(i^4)^q*i^r=1^q*i^r=i^r.
Also,note that 100,1000,10000 and every power of 10 bigger than 1,is a multiple of 4.Also,note that every number can be writen in the form: a0+a1*10+a2*100+a3*1000+a4*10000+...
and you will obtain it...

P.S.3.Anyway,if you want the complete proof,IM me and I'll try to write it down for you

Answer 2

Recall that i^4=1. So i^127 = i^(124+3) = i^124*1^3= i^3.
Now i² = -1, so i^3 = -i. The answer is -i.

Answer 3

we have i^2=-1
i^4=1
therefore i^127=i^(124+3)
this implies i^127=1*(i^3)=-i

i^127= - i

Answer 4

i^2 = -1 so
i^4 = 1
i^127 = (i^4)^31 * i^3
= 1 * i^2 * i
= -i

Answer 5

remember i^2 = -1, i^3 = -i, and i^4 = 1

so i^127 = i^100 * i^27
= (i^4)^25 * (i^3)^9
= 1 * (-i^9)
= 1*(-i^3)^3
= 1 * i^3
= 1*-i
= -i

Answer 6

i^(127) = i x i ^(126) = i x (- 1) = - i

Answer 7

i=-1
i^127=-i

Answer 8

i to an even power is 1 (e.g. i^2 = 1)
i to an odd power is -1 (e.g. i^3 = -1)

Therefore i^127 = -1

Answer 9

i is the answer

Answer 10

It's simply

i * i * i * i .........................* i (127 times)