Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
3  
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 $-y$

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

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

$x_2=-x_0$

$x_3=-x_1$

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

share|improve this answer

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.

share|improve this answer

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 $k$.

share|improve this answer

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

share|improve this answer

Hint: Use Complex numbers!!! In other words, to solve $x^4 + 1 = 0$, you have to find the four roots of unity.

share|improve this answer
1  
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! –  Learner 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 $(x^{2}+1i)(x^{2}-1i)$ $$(x+1\sqrt{i})(x-1\sqrt{i})(x+1i^{3/2})(x-1i^{3/2}) = x^{4}+1$$

share|improve this answer
2  
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
2  
Until you define precisely what the square root of i is, these are empty words. –  Did Nov 10 '12 at 18:29

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.