Question

How do you solve `u^4 = 1`?

Answer 1

`u=1,-1,i, -i`

Explanation:

Our equation is essentially a difference of squares which can be rewritten as

`(u^2+1)(u^2-1)`

To solve, we can set both of these equal to zero to get

`u^2=-1=>u=sqrt(-1)=+-i`

`u^2=1=>u=+-1`

Therefore, our solutions are

`u=1,-1,i, -i`

Hope this helps!

Answer 2

Given `4`th degree equation will have four roots : `\pm1, \pmi`

Explanation:

Given equation: `u^4=1` is a `4`th degree equation hence it will have `4` roots as follows

`u^4=1`

`u^4=e^{i(0)}`

`u^4=e^{i(2k\pi)}`

`u=(e^{i2k\pi})^{1/4}`

`u=e^{i1/4\cdot 2k\pi}`

`u=e^{i{k\pi}/2}`

`u=\cos({k\pi}/2)+i\sin({k\pi}/2)`

Where, `k` is an integer such that `k=0, 1, 2, 3`

Now, setting the values of `k` ,as `k=0, 1, 2 , 3` in above general solution, we get all four roots of given `4`th degree equation as follows

`u_1=\cos({(0)\pi}/2)+i\sin({(0)\pi}/2)=1`

`u_2=\cos({(1)\pi}/2)+i\sin({(1)\pi}/2)=i`

`u_3=\cos({(2)\pi}/2)+i\sin({(2)\pi}/2)=-1`

`u_4=\cos({(3)\pi}/2)+i\sin({(3)\pi}/2)=-i`

hence, all four roots are `\pm1, \pmi`

Answer 3

Answer below

Explanation:

`u^4 -1 = 0`

`=> (u^2+1)(u^2-1) = 0`

`=> u^2-1 = 0 => u = { -1 , 1 }`

`=> u^2 + 1= 0 => u = { i , - i }`

`therefore u = { -1 , 1 , -i , i }`