solve i^49?

Answer 1

i

i^1 = i
i^2 = -1
i^3 = -i
i^4 = 1

So, i^49 is i^48 (which is i^4 multiplied by itself 12 times) times i.

So i^49 = i.

Answer 2

i^48 = 1

so i^49 = i

Answer 3

i is the square root of -1, so i^2 = -1 and thus i^4 = 1.
i^49 = i * i^48 = i
since i^48 = 1^12 = 1.

Answer 4

You know that:

i^1 = i
i^2 = -1
i^3 = -i
i^4 = 1
and it continues in that pattern, so i^5=i, i^6=-1, etc

Now, i^49 can be written as (i^48)*(i^1)
Since 48 is a multiple of 4, i^48 = 1,
so (i^48)*(i^1) = 1*(i^1) = i

i^49 = i

Answer 5

i^49 = i^48*i = i

Answer 6

i^49 = i

Answer 7

First, express 49 as 2(24) + 1

i^49 =

i^(2(24) + 1))

Use exponential properties to separate and solve.

i * i^(2(24))

i * (i^2)^24

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

i * (-1)^24

And we know (-1) to an even power is +1, so we have

i * (1)

or simply

i

Answer 8

i = sqrt(-1)
i^2 = -1.
i^48 = -1^24
-1^24 = 1
i^49 = 1i

Answer 9

Its a constant, there is no solving to be done.