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Find the values of $x$, where $x \in \Bbb C$, for which

$$x^4-1 =0$$

I can see that $x^4-1 = (x^2-1)(x^2+1)=0$

So one set of roots can be taken from $$x^2-1=0$$$$ \Rightarrow x=\pm1$$

However, for $$x^2+1=0 $$$$\Rightarrow x=\sqrt{-1}$$$$\Rightarrow x=i$$

So from where does the last given answer of $-i$ come? I thought $i=\sqrt{-1}$ and $i^3=-i$, so can you please explain in what way does $-i$ work as a solution?

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$(-i)^2 = (-1)^2i^2 = 1\cdot (-1) = -1$ – Daniel Fischer Sep 7 '13 at 17:43
up vote 2 down vote accepted

A quadratic equation over $\mathbb{C}$ has always two solutions. So in particular, $x^2-1=0$ has two solutions, $x=1$ and $x=-1$, as has $x^2+1=0$, with $x=i$ and $x=-i$. The solutions of $x^4-1=0$ are called the $4$-th roots of unity, and they lie on the circle in the complex plane.

Here is the computation again: $(x+i)(x-i)=x^2-i^2=x^2+1$.

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"A quadratic equation over a field has always two solutions" ?? – Jonathan Sep 7 '13 at 18:27
Oops, I wanted to say, a polynomial of degree $n$ over a field has at most $n$ roots. – Dietrich Burde Sep 7 '13 at 18:30
what about $x^2+1$ over $\mathbb{R}$? – Jonathan Sep 7 '13 at 18:31
@Jonathan He's saying at most! – Pedro Tamaroff Sep 7 '13 at 18:57
@PeterTamaroff: no, it was my fault. I have edited my answer. So thanks to Jonathan. – Dietrich Burde Sep 7 '13 at 19:02

$x^2=a^2\implies x=\pm a$

If $a=i, a^2=-1, x=\pm i$


$x^2+1=0\implies x^2=-1=i^2\implies x=\pm i$

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You can always use Euler's formula: $$ e^{ix} = \cos(x) + i\sin(x) $$ where $x = \frac{2\pi k}{n}$ and $0\leq k < n$.

So in your case $n = 4$ and $k = 0, 1, 2, 3$.

\begin{alignat*}{2} \exp\left(\frac{i\pi k}{2}\right) &= 1 &&\quad\text{when } k = 0\\ &= i &&\quad\text{when } k = 1\\ &= -1 &&\quad\text{when } k = 2\\ &= -i &&\quad\text{when } k = 3 \end{alignat*}

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You can always use Euler's formula to solve the problem posed to the OP, but not to answer the question posed by the OP... – The Chaz 2.0 Sep 7 '13 at 18:13

The bug in your work is:

$x^2+1=0 \implies x=\sqrt{-1}$.

A counterexample to this is $x=-\sqrt{-1}$ also satisfies $x^2+1=0$. (The implication actually goes the other way: $x=\sqrt{-1} \implies x^2+1=0$.)

The claim

$x^2-1=0 \implies x=\pm 1$

is implied by the Fundamental Theorem of Algebra. Specifically, the corresponding polynomial has exactly $2$ (not necessarily distinct) roots. Having found $2$ roots, we can stop looking.

Again, the Fundamental Theorem of Algebra implies that there are exactly $4$ (not necessarily distinct) roots of a degree $4$ polynomial; in this case $x^4-4$. This means that as soon as we've found $4$ roots, we can stop searching, as there are no more roots.

Now, for any polynomial $f$ with complex coefficients, if $z$ is a root, then the complex conjugate of $z$ is also a root; Wikipedia has a page on this: Complex Conjugate Root Theorem. In this case, $i$ is a root, and the theorem implies that $-i$ must also be a root.

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