# Find the values and roots of the equation z^4-2z^3+7z^2-4z+10=0?

## Find the values of $a \in \mathbb{R}$ for which $a i$ is a solution of ${z}^{4} - 2 {z}^{3} + 7 {z}^{2} - 4 z + 10 = 0$ Also find all the roots of this equation.

Mar 13, 2017

The roots are: $\pm \sqrt{2} i$ and $1 \pm 2 i$

#### Explanation:

Given:

${z}^{4} - 2 {z}^{3} + 7 {z}^{2} - 4 z + 10 = 0$

If $z = a i$ for some real number $a$ then, the terms of even degree are real and those of odd degree imaginary...

$0 = {\left(a i\right)}^{4} - 2 {\left(a i\right)}^{3} + 7 {\left(a i\right)}^{2} - 4 \left(a i\right) + 10$

$\textcolor{w h i t e}{0} = \left({a}^{4} - 7 {a}^{2} + 10\right) + 2 a \left({a}^{2} - 2\right) i$

$\textcolor{w h i t e}{0} = \left({a}^{2} - 5\right) \left({a}^{2} - 2\right) + 2 a \left({a}^{2} - 2\right) i$

$\textcolor{w h i t e}{0} = \left(\left({a}^{2} - 5\right) + 2 a i\right) \left({a}^{2} - 2\right)$

Hence $a = \pm \sqrt{2}$

So two of the roots of the original quartic are $\pm \sqrt{2} i$, with associated factors:

$\left(z - \sqrt{2} i\right) \left(z + \sqrt{2} i\right) = {z}^{2} + 2$

We find:

${z}^{4} - 2 {x}^{3} + 7 {z}^{2} - 4 z + 10 = \left({z}^{2} + 2\right) \left({z}^{2} - 2 z + 5\right)$

$\textcolor{w h i t e}{{z}^{4} - 2 {x}^{3} + 7 {z}^{2} - 4 z + 10} = \left({z}^{2} + 2\right) \left({z}^{2} - 2 z + 1 + 4\right)$

$\textcolor{w h i t e}{{z}^{4} - 2 {x}^{3} + 7 {z}^{2} - 4 z + 10} = \left({z}^{2} + 2\right) \left({\left(z - 1\right)}^{2} - {\left(2 i\right)}^{2}\right)$

$\textcolor{w h i t e}{{z}^{4} - 2 {x}^{3} + 7 {z}^{2} - 4 z + 10} = \left({z}^{2} + 2\right) \left(z - 1 - 2 i\right) \left(z - 1 + 2 i\right)$

So the other two roots are:

$z = 1 \pm 2 i$

Mar 13, 2017

$a = \setminus \pm \sqrt{2}$
$z = \setminus \pm \sqrt{2} i , 1 \setminus \pm \sqrt{2} i$

#### Explanation:

We can use the fact that the problem asks us to find values of $a$ so that $a i$ is a solution. Thus, if $a i$ is a solution, so is the conjugate, or $- a i$.

Let us denote the other two solutions as $c \setminus \pm \mathrm{di}$. If we write out our polynomial, say $P \left(z\right)$, by way of the roots, we have that:

$P \left(z\right) = \left(z - a i\right) \left(z + a i\right) \left(z - \left(c - \mathrm{di}\right)\right) \left(z - \left(c + \mathrm{di}\right)\right) = 0$.

We can expand this, group like terms, and find that:
$P \left(z\right) = {z}^{4} - \left(2 c\right) {z}^{3} + \left({c}^{2} + {d}^{2} + {a}^{2}\right) {z}^{2} - \left(2 {a}^{2} c\right) z + \left({a}^{2} {c}^{2} + {a}^{2} {d}^{2}\right) = 0$.

Set like-coefficients equal (so set $- 2 {z}^{3} = - 2 c$ etc), we find that:

$a = \setminus \pm \sqrt{2}$
$z = \setminus \pm \sqrt{2} i , 1 \setminus \pm \sqrt{2} i$

By graphing, we see that this solution set works.