# How do you find the complex roots of t^4-1=0?

Jan 7, 2017

${t}^{4} - 1 = 0$ has roots $\pm 1$ and $\pm i$

In a sense these are all complex roots, but $\pm i$ particularly so, in that they are not Real roots.

#### Explanation:

The difference of squares identity can be written:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

We find:

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

$\textcolor{w h i t e}{0} = {\left({t}^{2}\right)}^{2} - {1}^{2}$

$\textcolor{w h i t e}{0} = \left({t}^{2} - 1\right) \left({t}^{2} + 1\right)$

$\textcolor{w h i t e}{0} = \left({t}^{2} - {1}^{2}\right) \left({t}^{2} - {i}^{2}\right)$

$\textcolor{w h i t e}{0} = \left(t - 1\right) \left(t + 1\right) \left(t - i\right) \left(t + i\right)$

Hence $t = \pm 1$ or $t = \pm i$

The roots $\pm i$ are non-Real Complex roots.

The roots $\pm 1$ are Real roots. Since Real numbers are a subset of Complex numbers, $\pm 1$ are also Complex roots.