# How do you find all the real and complex roots of z^7+z^4+4z^3+4=0?

Jan 19, 2016

$z = - 1 , - 1 \pm i , 1 \pm i , \frac{1}{2} \pm \frac{\sqrt{3}}{2} i$

#### Explanation:

Factor by grouping:

${z}^{4} \left({z}^{3} + 1\right) + 4 \left({z}^{3} + 1\right) = 0$

$\left({z}^{4} + 4\right) \left({z}^{3} + 1\right) = 0$

First, $\left({z}^{3} + 1\right)$ can be easily split apart using the sum of cubes identity.

$\left({z}^{4} + 4\right) \left(z + 1\right) \left({z}^{2} - z + 1\right) = 0$

The other part, $\left({z}^{4} + 4\right)$, is more difficult to factor. We can try to turn it into a difference of squares.

${z}^{4} + 4 = \left({z}^{4} + 4 {z}^{2} + 4\right) - 4 {z}^{2}$
$\textcolor{w h i t e}{\times \times} = {\left({z}^{2} + 2\right)}^{2} - {\left(2 z\right)}^{2}$
$\textcolor{w h i t e}{\times \times} = \left({z}^{2} + 2 z + 2\right) \left({z}^{2} - 2 z + 2\right)$

Now, we have $3$ quadratic factors and one linear factor:

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

Solving each of these via the quadratic formula or completing the square gives the seven distinct values of $z$:

$z = - 1 , - 1 \pm i , 1 \pm i , \frac{1}{2} \pm \frac{\sqrt{3}}{2} i$