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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.


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Can you find the roots of $x^2+1$? Once you do that, take a square root of each! –  Lubin Nov 10 '12 at 5:29

6 Answers 6

up vote 2 down vote accepted

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 $-y$

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

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



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

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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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More precisely, the four primitive eighth roots of unity! –  Lubin Nov 10 '12 at 5:28
nice proof $i^i$ is real!! Very very cute! –  Marvin Gaye Nov 10 '12 at 5:30

Well, you could use $$(x+1)(x-1) = x^{2}-1$$ Use the same property for positive $$(x+1i)(x-1i) = x^{2}+1$$ Then $$(x+1)(x-1)(x+1i)(x-1i) = x^{4}-1$$ To make it positive just add the $\sqrt{i}$ to each factor in order to obtain $(x^{2}+1i)(x^{2}-1i)$ $$(x+1\sqrt{i})(x-1\sqrt{i})(x+1i^{3/2})(x-1i^{3/2}) = x^{4}+1$$

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What do you call $\sqrt{i}$ and $i^{3/2}$? –  Did Nov 10 '12 at 7:18
Well, i is the imaginary number hich is $i = \sqrt{-1}$. They are complex numbers. –  Fabián H. jr. Nov 10 '12 at 15:07
This was not my question. –  Did Nov 10 '12 at 15:50
The square root of i and i to the 1 and a half power? –  Fabián H. jr. Nov 10 '12 at 16:04
Until you define precisely what the square root of i is, these are empty words. –  Did Nov 10 '12 at 18:29

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