i is square root of -1. Is it Clear. Iam asking Whether i power n is always 1 or not.?

Question

Refer to my Previous question For any Doubt. Is it Clear.

Answer 1

In addition to the points already given, the best way to solve for
i^n is to express n in the form 2k + 1. I'll show you some examples:

1) i^24

Note that 24 = 2(12). Therefore,

i^24 = i^[2*12]

Use the property of exponents that allows you to express
a^(bc) as (a^b)^c.

i^24 = (i^2)^12

But, we know that i^2 = -1, so

i^24 = (-1)^12

And we also know that (-1) to an even power is 1, so

i^24 = 1

Example #2:

i^33

31 = 2(15) + 1, so

i^33 = i^[2(15) + 1]

Use the property a^(b + c) = (a^b)(a^c).

i^33 = [ i^(2*15) i^1 ]

But i^1 = 1, so

i^33 = [ i^(2*15) i ]

As done in the previous question,

i^33 = [ (i^2)^(15) i ]
i^33 = [ (-1)^15 i ]

-1 is to an odd power, so it is just equal to -1.

i^33 = (-1) i

And so the final answer is -i.

3) i^997 = i^(2*498 + 1)
= i^(2*498) i
= (i^2)^498 * i
= (-1)^498 * i
= (1) i
= i

Answer 2

No, i has a four number cycle that it follow for being raised to whole number powers:

i^1=i (n^1 always equals n)
i^2=-1 (i is the sqrt of -1, so i^2 is equal to -1)
i^3=-i (think of it as i^1 x i^2, which is i x -1, equalling -i)
i^4=1 (similiar to the last, think of it as i^2 x i^2, -1x-1=1)
i^5=i (i^1 x i^4, i x 1 equals i)

Now you're back where you started, so just keep following the pattern.

Answer 3

square root is not applicable to -ve integers, but OF COURSE 1 power n or 1 power 0 is always 1

Answer 4

i = sqrt (-1) i.e. i square = -1, i cube = -i, i power 4 = 1. just keep multiplying remembering that i square = -1. That means i power n is not always 1. It could be i, -i, -1 or 1.

Answer 5

no thats not clear but the square root of 1 is 1 isnt it

Answer 6

i to power n (usually written i^n here) is
-1 if n=2,
but if n = 3 then clearly
i*i*i
= -1*i
= -i
Then of course if n = 4, ...
OK I'll stop there, as I see Daniel's already done it.
I just add that
i^n = 1if n is a multiple of 4 (i.e. 4p for integer p)
i if n is 1 more than a multiple of 4 (i.e. 4p+1)
-1 if n = 4p + 2
-i if n = 4p+3

Answer 7

i square is -1 n i is square root -1
n i power n is or not =1 depends upon n
if n=2 then i2= -1, i3= -i, i4= -1*-1=1,i5= i(3+2)= -i*-1=i n so on...
(refer i2 as i power 2 n so on)

Answer 8

WE KNOW THAT SQUAR ROOT OF –1 IS “i” AND AS A RULE IN COMPLEX NUMBERS, VALUE OF “i” POWER n DEPENDS ON EITHER THE n IS AN ODD OR EVEN NUMBER, IF IT IS ODD THE RESULT WILL BE –1 OTHERVISE IT WILL BE 1.

Answer 9

i or j sometimes it is called is used to represent complex coordinates .Coming to your question of i to the power of n the answer is 1 if n is even
-1 if n is odd.

Answer 10

i is d square root of -1...true....but its not true dat i^n is always 1.....
for example i^2 is -1...