What is the fourth root of i to the the forth power?

Question

i^1=(-1)^(1/2)
i^2=(-1)^((1/2)*2)=-1
i^3=(-1)^((1/2)*3)=((-1)^((1/2)*2))*(-1)^(1/2)=(-1)(i)=-i
i^4=(-1)^((1/2)*4)=((-1)^((1/2)*2))*((-1)^((1/2)*2))=(-1)(-1)=1

i^4=1 => (i^4)^(1/4)=(i^(4*(1/4)))= i = 1

i =?= 1

Answer 1

i to the fourth power is 1.

So your question is asking what is (are) the fourth root(s) of 1. In other words, what are the complex numbers whose fourth power is 1. There are four answers:

1, -1, i, -i.

Those four complex numbers raised to their fourth power all give you 1.

Answer 2

If you are talking about the general function of an even root, the result is defined to be real and positive for real positive arguments, or imaginary and positive for real negative arguments. So, the fourth root of i^4 is 1. However, if you mean the general solution to the equation x^(1/4) = C, there will be multiple results in the complex plane. For the equation 1^(1/4), the roots are +/- 1 and +/- i.

Actually, there are always multiple results for any root operation. For a fourth root, there are four possible answers; for an Nth root, there are N possible answers. They are evenly spaced in a circle on the complex plane. If N is even, two roots will fall on the real axis: one positive, and one negative. For example, sqrt(4) = +/- 2. If N is odd, only one root will fall on the real axis; 8^(1/3) = 2. However, the rest of the answers still exist. For example,

8^(1/3) = 2@0°, 2@120°, 2@240°
= 2cos0° + 2sin0°, 2cos120° + 2sin120°, 2cos240° + 2sin240°
= 2, -1 + i sqrt(3), -1 - i sqrt(3)
So the cube roots of 8 are:

2
-1 + i sqrt(3)
-1 - i sqrt(3)

Answer 3

Actually there are *four* 4th roots to the expression i^4 (a.k.a. 1). They are 1, -1, i and -i, as shown below:

1 x 1 x 1 x 1 = 1
-1 x -1 x -1 x -1 = 1
i x i x i x i = 1
-i x -i x -i x -i = 1

But none of these roots (1, -1, i, -i) are equivalent.

This is similar to taking the square root of 4. There are two roots (-2 and 2)... but we wouldn't say -2 equals 2, would we?

Answer 4

Any (x^4)^1/4 is equivalent to x^4/4 = x^1 = x

So the fourth root of i to the the fourth power is i.

Answer 5

Fourth root of i = i^(1/4)
...to the 4th power = [i^(1/4)]^4 = i^[(1/4)4] = i

Answer 6

fourth root of (i to the the forth power) = fourth root of 1 = 1
(This is an other question than solve x^4 = 1)

Th

Answer 7

i^(1/4))^4=i^(4/4)=i^1=i

Answer 8

the 4th root of any number x is x^(1/4).

Any number x raised to the 4th power is x^4

Thus your number (x^1/4)^4 = x^(4/4) = x^1

Substitute i for x and you get i^1 which = i

Answer 9

fourth root of i to the fourth power is i itself
(i^4)^1/4
=i

Answer 10

the fourth root of i to the fourth power is simply i

fourth root and power of four cancel each other out.