# How do you solve u^4 = 1?

Jul 21, 2018

$u = 1 , - 1 , i , - i$

#### Explanation:

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

$\left({u}^{2} + 1\right) \left({u}^{2} - 1\right)$

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

${u}^{2} = - 1 \implies u = \sqrt{- 1} = \pm i$

${u}^{2} = 1 \implies u = \pm 1$

Therefore, our solutions are

$u = 1 , - 1 , i , - i$

Hope this helps!

Given $4$th degree equation will have four roots : $\setminus \pm 1 , \setminus \pm i$

#### Explanation:

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

${u}^{4} = 1$

${u}^{4} = {e}^{i \left(0\right)}$

${u}^{4} = {e}^{i \left(2 k \setminus \pi\right)}$

$u = {\left({e}^{i 2 k \setminus \pi}\right)}^{\frac{1}{4}}$

$u = {e}^{i \frac{1}{4} \setminus \cdot 2 k \setminus \pi}$

$u = {e}^{i \frac{k \setminus \pi}{2}}$

$u = \setminus \cos \left(\frac{k \setminus \pi}{2}\right) + i \setminus \sin \left(\frac{k \setminus \pi}{2}\right)$

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

${u}_{1} = \setminus \cos \left(\frac{\left(0\right) \setminus \pi}{2}\right) + i \setminus \sin \left(\frac{\left(0\right) \setminus \pi}{2}\right) = 1$

${u}_{2} = \setminus \cos \left(\frac{\left(1\right) \setminus \pi}{2}\right) + i \setminus \sin \left(\frac{\left(1\right) \setminus \pi}{2}\right) = i$

${u}_{3} = \setminus \cos \left(\frac{\left(2\right) \setminus \pi}{2}\right) + i \setminus \sin \left(\frac{\left(2\right) \setminus \pi}{2}\right) = - 1$

${u}_{4} = \setminus \cos \left(\frac{\left(3\right) \setminus \pi}{2}\right) + i \setminus \sin \left(\frac{\left(3\right) \setminus \pi}{2}\right) = - i$

hence, all four roots are $\setminus \pm 1 , \setminus \pm i$

Jul 21, 2018

#### Explanation:

${u}^{4} - 1 = 0$

$\implies \left({u}^{2} + 1\right) \left({u}^{2} - 1\right) = 0$

$\implies {u}^{2} - 1 = 0 \implies u = \left\{- 1 , 1\right\}$

$\implies {u}^{2} + 1 = 0 \implies u = \left\{i , - i\right\}$

$\therefore u = \left\{- 1 , 1 , - i , i\right\}$