# Sum of non-real roots of equation?

What is the sum of all non-real, complex roots of this equation -

$$x^5 = 1024$$

Also, please provide explanation about how to find sum all of non real, complex roots of any $n$ degree polynomial. Is there any way to determine number of real and non-real roots of an equation?

• The sum is $0$. – Lucian Aug 7 '15 at 5:14
• Can you please explain how you got that answer? – Confuse Aug 7 '15 at 5:14
• @Lucian OP asked for the sum of non-real, complex roots – user217285 Aug 7 '15 at 5:18
• @Nitin: Either way, the answer should be obvious. Just subtract the obvious real root from the sum of all $5$. – Lucian Aug 7 '15 at 5:27
• The only real root is $x=4$ because over the reals $x^5$ is an increasing function. – André Nicolas Aug 7 '15 at 5:45

By the fundamental theorem of algebra, the polynomial $a_0 + a_1x + \cdots + a_nx^n$ has $n$ (not necessarily distinct) roots, say $\alpha_1,\ldots,\alpha_n$. The polynomial $\frac{a_0}{a_n} + \frac{a_1}{a_n}x + \cdots + x^n$ also has roots $\alpha_1,\ldots,\alpha_n$. Factoring the latter polynomial and expanding, \begin{align*} \frac{a_0}{a_n} + \frac{a_1}{a_n}x + \cdots + \frac{a_{n-1}}{a_n} x^{n-1} + x^n &= (x - \alpha_1)\cdots(x - \alpha_n) \\ &= x^n - (\alpha_1 + \cdots+ \alpha_n)x^{n-1} + \cdots. \end{align*} Comparing coefficients, we have $\alpha_1 + \cdots + \alpha_n = -\frac{a_{n-1}}{a_n}$, giving us an explicit formula for the sum of the roots.

We are looking for the sum of the non-real roots of $x^5 - 1024$. Clearly $x = 4$ is a root, and since $5$ is odd, there are no other real roots (in the complex plane, the roots lie on a circle with radius 4 centered at the origin in the shape of a pentagon with a vertex at $(4,0)$). The sum of all the roots is 0, so the sum of the non-real roots is $-4$.

• What if there were 3 or 5 real roots? How can you say that there are no more real roots? – Arulx Z Aug 10 '15 at 6:48
• The geometric argument (pentagon in the complex plane) suffices here. I used this approach over the more standard "$x^5$ is increasing" argument because the question was explicitly asking about roots. – user217285 Aug 10 '15 at 6:49

Hint: By Vieta, sum of all roots is $0$, and the only real root is $4$.

P.S. For a simple argument that there is only one real root, apply Descartes' rule of signs on $x^5-1024$.

• so the sum of the complex roots is $-4$... – hjhjhj57 Aug 7 '15 at 5:25
• @hjhjhj57 Yes... – Macavity Aug 7 '15 at 5:25

$\bf{My\; Solution::}$ Given $$x^5 = 1024 = 2^{10} = 4^5\Rightarrow x^5-4^5 = 0$$

So $$(x^5-4^5) = (x-5)\cdot (x^4+x^3\cdot 4+x^2\cdot 4^2+x\cdot 4^3+4^4) = 0$$

Above we use the formula $\displaystyle x^n-a^n = (x-a)\cdot (x^{n-1}+x^{n-2}\cdot a+x^{n-3}\cdot a^2+.......+a^{n-1})$

So We Get $x=4(\bf{Real \; Root})$ and $$x^4+x^3\cdot 4+x^2\cdot 4^2+x\cdot 4^3+4^4=0$$

So Sum of Roots in $\bf{Second}$ Equation is $\displaystyle = -\frac{4}{1} = -4$

Above we use the Formula $\displaystyle \bf{Sum\; of \; Roots} = -\frac{\bf{coeff.\; of \; x^3}}{\bf{coeff.\; of \; x^4}}$

• How do you know, that roots of the residual polynomial are all complex? – Kaster Aug 7 '15 at 5:22
• Since $x^{5}=1024$ implies that $x=\sqrt[5]{1024}=4=x_0$ is one of the roots of $x^{5}-1024=0$, it follows that $$p(x)=x^{5}-1024=(x-4)q(x),$$ where $q(x)$ is a forth degree polynomial with real coefficients and leading term equal to $x^{4}$. Hence it is of the form $$q(x)=x^{4}+bx^{3}+cx^{2}+dx+e.$$
• The coefficients $b,c,d, e$ can be found by polynomial long division (see below1) or by Ruffini's Rule. Using the latter to compute $$\frac{p (x)}{x-{\color{blue}4}}=q (x),$$ i.e. \begin{array}{c|rrrrrl} & x^{5} & x^{4} & x^{3} & x^{2} & x^{1} & \phantom{-} x^{0} \\ & 1 & 0 & 0 & 0 & 0 & -1024\\ x_0= {\color{blue}4} & \downarrow & 4 & 16 & 64 & 256 &\phantom{-} 1024\\ \hline & 1 & 4 & 16 & 64 & 256 & \phantom{-102} {\color{blue}0}=\text{remainder}\\ & x^{4} & x^{3} & x^{2} & x^{1} & x^{0} \end{array} yields \begin{equation*} q(x)=x^{4}+4x^{3}+16x^{2}+64x+256. \end{equation*} The polynomial $q(x)$ has four roots we denote as $x_{1},x_{2},x_{3},x_{4}$. Since the coefficient of $x^{4}$ is $1$, we know that it can be written as \begin{eqnarray*} q(x) &=&\left( x-x_{1}\right) \left( x-x_{2}\right) \left( x-x_{3}\right) \left( x-x_{4}\right) \\ &=&x^{4}-\left( x_{1}+x_{2}+x_{3}+x_{4}\right) x^{3} \\ &&+\left( x_{3}x_{4}+x_{2}x_{4}+x_{2}x_{3}+x_{1}x_{4}+x_{1}x_{3}+x_{1}x_{2}\right) x^{2} \\ &&-\left( x_{1}x_{2}x_{3}+x_{1}x_{2}x_{4}+x_{1}x_{3}x_{4}+x_{2}x_{3}x_{4}\right) x+x_{1}x_{2}x_{3}x_{4} \end{eqnarray*}
• Now by comparing coefficients of $x^{3}$ we see that \begin{equation*} -b=x_{1}+x_{2}+x_{3}+x_{4}=-4, \end{equation*} which is just Vieta's formula for the sum of the roots of the polynomial $q (x)$. The original equation $p(x)=0$ has thus five roots: $x_0=4$ and the four roots $x_{1},x_{2},x_{3},x_{4}$ of $q(x)$. We just need to prove that none of these $x_{1},x_{2},x_{3},x_{4}$ are real numbers.

For this purpose we can use simple Calculus. Since $p'(x)=5x^4>0$ for $x\neq 0$ and $p'(0)=0$, the polynomial $p(x)$ is an increasing function and

1. if $x< 4$, then $p(x)<0$,
2. if $x>4$, then $p(x)>0$,
3. $p(4)=0$,

all roots of $q(x)=0$ are non-real complex numbers, whose sum as computed above equals $-4$.

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1 Polynomial long division $p (x)/(x-{\color{blue}4})=q(x)$

\begin{array}{rl} &\underline{\phantom{+x^{5}} \phantom{+0}x^{4}\phantom{0}+4x^{3}+16x^{2}\phantom{0}+64x\phantom{0}+256\phantom{0}} \\ x-{\color{blue}4}) & \phantom{+}x^{5}\phantom{+0x^{4}0+4x^{3}+16x^{2}+064x}-1024 \\ & \underline{-x^{5}+4x^4\phantom{0}} \\ & \phantom{-x^{5}+}\;4x^4 \\ & \phantom{-x^{5}}\underline{\;-4x^4+16x^3\phantom{0}} \\ & \phantom{-x^{5}-4x^4+}16x^3 \\ & \phantom{-x^{5}-4x^4}\underline{\;-16x^3+64x^{2}\phantom{0}} \\ & \phantom{-x^{5}-4x^4+16x^3+}64x^{2}\\ & \phantom{-x^{5}-4x^4+16x^3}\underline{\;-64x^{2}+256x\phantom{0}} \\ & \phantom{-x^{5}-4x^4+16x^3-64x^{2}+}256x-1024 \\ & \phantom{-x^{5}-4x^4+16x^3-64x^{2}}\underline{\;-256x+1024\phantom{0}} \\ & \phantom{-x^{5}-4x^4+16x^3-64x^{2}-256x+102}{\color{blue}0} \end{array}