Find the values of $x$, where $x \in \Bbb C$, for which $x^4-1 =0$ 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? 
 A: 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: $(x+i)(x-i)=x^2-i^2=x^2+1$.
A: $x^2=a^2\implies x=\pm a$
If $a=i, a^2=-1, x=\pm i$
i.e.,
$x^2+1=0\implies x^2=-1=i^2\implies x=\pm i$
A: 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*}
A: 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.
