Math help plz!?

Question

How do I:

Simplify the complex number i^58 as much as possible?

Thanks so much for your help!

Answer 1

The best way is to make use of the property a^(bc) = (a^b)^c.

Let's use the fact that 58 = 2*29.

i^58 = i^(2*29)
= (i^2)^29

Note that i^2, by definition, is equal to (-1). Therefore, we get

(-1)^(29)

And (-1) to an odd power is itself, so our answer is just

(-1)

***
Here are some more miscellaneous tips regarding i.

For odd powers of i (for instance 129), perform these steps.

i^(129)

1) Break off an i, to make an even power.

i * i^(128)

2) Express the now even power as a multiple of 2.

i * i^(2*64)

3) Use the exponential property to change the term into a power to a power.

i * (i^2)^(64)

i^2 = -1, so

i * (-1)^(64)

(-1) to an even power is 1.

i * 1^(64)

1 to the power of anything is 1.

i * 1

i

Answer 2

That would be 1.

Answer 3

i^0=1
i^1=i
i^2=-1
i^3=-ì
and i^4=1 so you have a repetition after a power of 4
Divide 58 by 4 and take the reamainder
i^58 = (i^4)^14 *i^2 = -1 as the first factor is 1

Answer 4

Subtract multiples of 4. i^58 = i^56(i^2) = 1(i^2) = -1

Answer 5

Well i^4 = 1, since i squared is -1, right?

So i^56 is also 1, since 4 divides into 56. So you have two i's left, so to speak, hence the answer is i^2 which is -1

Answer 6

-1 i