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I'm trying to find the roots of $x^4+1$. I've already found in this site solutions for polynomials like this $x^n+a$, where $a$ is a negative term. I don't remember how to solve an equation when $a$ is a positive term as the equation above.

Thanks

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    $\begingroup$ Can you find the roots of $x^2+1$? Once you do that, take a square root of each! $\endgroup$
    – Lubin
    Nov 10, 2012 at 5:29

8 Answers 8

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Another way is to use some creative rewriting:

$$x^4 + 1 = (x^2 + 1)^2 - 2x^2 = (x^2 + 1 - x\sqrt 2)(x^2 + 1 + x\sqrt2).$$

Then just solve the two quadratic equations.

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    $\begingroup$ Genius, didn't think about this trick xD $\endgroup$
    – gedamial
    Jul 6, 2019 at 13:41
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Using this, $x^4=-1=e^{i\pi}=e^{(2n+1)\pi i}$ as $e^{2m\pi i}=1$ where $m,n$ are integers.

So, $x=e^{\frac{(2n+1)\pi i}4}=\cos\frac{(2n+1)\pi}4 +i\sin \frac{(2n+1)\pi}4 $ where $n$ has any $4$ in-congruent values $\pmod 4$, the most simple set of values of $n$ can be $\{0,1,2,3\}$.

If $x_m=e^{\frac{(2m+1)\pi i}4},x_{m+2}=e^{\frac{(2m+3)\pi i}4}=e^{\frac{(2m+1)\pi i}4}\cdot e^{\frac {i\pi}2}=-x_m$

Also, observe that if $y$ is a solution of $x^4=-1$, so is $-y$

$x_0=\cos\frac{\pi}4 +i\sin \frac{\pi}4=\frac{1+i}{\sqrt 2}$

$x_1=\cos\frac{3\pi}4 +i\sin \frac{3\pi}4=\frac{-1+i}{\sqrt 2}$

$x_2=-x_0$

$x_3=-x_1$

So, the values of $x$ are $\pm\left(\frac{1+i}{\sqrt 2}\right),\pm\left(\frac{-1+i}{\sqrt 2}\right)$

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$$ x^4=-1 $$ $$ x^2=\pm i = \pm\left(\cos\frac\pi2 + i\sin\frac\pi2\right) $$ $$ \text{If }x^2 = \cos\frac\pi2 + i\sin\frac\pi2 \text{ then } x = \pm\left(\cos\frac\pi4+i\sin\frac\pi4\right). $$ $$ \text{If }x^2 = -\left(\cos\frac\pi2 + i\sin\frac\pi2\right) = \cos\frac\pi2 - i\sin\frac\pi2\text{ (since $\cos\frac\pi2=0$)} $$ $$ \text{then }x=\pm\left(\cos\frac\pi4-i\sin\frac\pi4\right). $$

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Because $x^8-1=(x^4+1)(x^4-1)$, the roots of $x^4+1$ are the roots of $x^8-1$ that are not roots of $x^4-1$.

The roots of $x^n-1$ are easy: $e^{\frac{2k\pi i}{n}}$ for $k=0,\dots,n-1$.

Therefore, roots of $x^4+1$ are $e^{\frac{2k\pi i}{8}}$ for $k$ odd.

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Finding the roots of $X^n=z$ in $\mathbb{C}$ is a simple problem if you use the exponential notation for $z$.

If $z=\rho e^{i\theta}$ then the $n$ roots of this polynomial are: $\{\rho^{\frac{1}{n}}e^{i\frac{\theta+2k\pi}{n}},k\in [\![ 0;n-1 ]\!] \}$.

In this case, $n=4$, $\rho=1$ and $\theta=\pi$, so the roots are $e^{i\pi\frac{2k+1}{4}}$ for $k\in[\![ 0;3 ]\!]$,

which can also be written $e^{i\frac{\pi}{4}}i^k$ for the same values of $k$.

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We have:

$$\begin{align}x^4+1 =0 \\ x^4+1+2x^2-2x^2 =0 \\ (x^2+1)^2-(\sqrt{2}x)^2 =0 \\ (x^2+1+\sqrt{2}x)(x^2+1-\sqrt{2}x) =0 \end{align}$$

Then on solving these factors by quadratic formula, we will have:

$$x= \frac{-1}{\sqrt{2}}+\frac{i}{\sqrt{2}} , \frac{-1}{\sqrt{2}}-\frac{i}{\sqrt{2}}, \frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}, \frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}}$$

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    $\begingroup$ Welcome to MSE. Please format using MathJax $\endgroup$
    – user507623
    Dec 19, 2017 at 19:54
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Hint: Use Complex numbers!!! In other words, to solve $x^4 + 1 = 0$, you have to find the four roots of unity.

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    $\begingroup$ More precisely, the four primitive eighth roots of unity! $\endgroup$
    – Lubin
    Nov 10, 2012 at 5:28
  • $\begingroup$ nice proof $i^i$ is real!! Very very cute! $\endgroup$
    – ILoveMath
    Nov 10, 2012 at 5:30
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There is a very simple and clear way to solve this using exponentials. Here is the copy of an answer I wrote in a duplicate question yesterday:

We know that $1 = \exp{(2ik\pi})$ and $-1 = \exp{(2ik\pi+\pi})$ with $k \in \mathbb{Z}$, so why not writing:

$$\rho\exp{(4i\theta)} = \exp{(2ki\pi+\pi)}$$ $$\Longleftrightarrow > \begin{cases}\theta = k\pi/2+\pi/4\\ \rho=1 \end{cases}$$

So the solutions would be:

$$z_k = \exp\left(\frac{ik\pi}{2} + \frac{i\pi}{4}\right)$$

You can very easily extrapolate this to any value of $a$.

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