Motivational example for complex numbers Years ago I was introduced to complex numbers. In class we had been talking about the cubic polynomial and its solutions. At one point we saw an example where, when using the formula, one had to stop because one had to take a square root of a negative number. The teacher went ahead and expressed things in terms of the square root of negative 1. The surprise was that the solutions that we got in the end were real solutions.
The point of the example was that if we just ignore the square root of a negative number and continue using the algorithm/formula for the solution, then everything actually worked out.
My question/request is to see such an example redone. I haven't been able to find an example like that.
 A: You might be interested in looking at Cardano's method for solving cubics. Cardano himself didn't use complex numbers, but we can. Here's what happens when we use his method to solve, say, $x^3-x=0$.
First set $x=u+v$, then rearrange the result so that we have:
$$u^3 + v^3 + (3uv-1)(u+v) = 0.$$
Now assume that $3uv=1$. This implies from the above equation that $u^3+v^3=0$, and it also means that $u^3v^3=\frac{1}{27}$. We now have two equations relating the sum and product $u^3$ and $v^3$, which can be solved using the quadratic formula.
In particular, we will get $u^3 = \sqrt{-\frac{1}{27}}$, at which point Cardano would have chosen a different method. But if we forge on using complex numbers, we'll get three choices for $u$:
$$u_1 = -i\frac{\sqrt3}{3},$$
$$u_2 = u_1\big(-\frac{1}{2}+\frac{\sqrt3}{2}i\big),$$
$$u_3 = u_1\big(-\frac{1}{2}-\frac{\sqrt3}{2}i\big).$$
(That is, after finding $u_1$, the other two are rotations of it by $\pm120^\circ$.) It is straightforward to find the corresponding values for $v$. Finally, coming back to $x=u+v$, all of the imaginary components will happily cancel out. For instance,
$$x_1 = u_1 + v_1 = -i\frac{\sqrt3}{3} + i\frac{\sqrt3}{3} = 0.$$
We find that the three roots are $0$, $1$, and $-1$.
A: Théophile's answer is the standard one, but I'd like to offer a thought related to your question, if you're familiar with first-year Calculus (or even if you remember Geometric Series if you took high school precalculus). This answer, then, is more meant to supplement Théophile's. 
The following two "infinitely long" series of sums is valid for all real values of $\left|x\right| < 1$, using the Geometric Series formula (or calculating Taylor series at $x=0$, if you're familiar with Calculus): 
$$f(x)=\frac{1}{1-(-x)}=1-x+x^2-x^3+...$$
$$g(x)=\frac{1}{1-(-x^2)}=1-x^2+x^4-x^6+...$$
We might intuit that the geometric series for $f(x)$ is valid only up to $\left|x\right|<1$ because when $|x|=1$ - that is, either if $x=1$ or $x=-1$ - we have a problem, namely if we tried to "plug in" $-1$ into $f(x)$, we would get $$\frac{1}{1-1}=\frac{1}{0}$$Alas, we can't divide by zero, so we conclude $x=-1$ is not in the domain of $f(x)$. So, in a sense, the values of $x$ for which the geometric series representation for $f(x)$ can be assured to be valid must be those values whose distance from $0$ is less than $1$. Intuitively, then, we can get a general intuition about where series of the types listed above work for sure - they work until, at some distance from zero, there's a "snag" somewhere that makes us less sure they could work at that distance or any further out. 
But this thinking fails for $g(x)$. Notice, since $x^2>0$, the denominator $1-(-x^2)=1+x^2 \ge 1 >0$, so there is no value of $x$ that forces division by zero. This makes the rule that $|x|<1$ intuitively mysterious. With our thinking for $f(x)$, there ought to be a value at a distance of $1$ from $0$ (namely, either the numbers $1$ or $-1$) that would force us to have to divide by zero. But $g(1)=g(-1)=\frac{1}{2}$. So why do we have to stop using this geometric series that far away from zero? 
This is where complex numbers come along to save us. Since we are figuring out why our geometric series is only guaranteed to work for $|x|<1$, we want to find a number that forces us to "divide by zero" in $g(x)$. That would mean, now allowing complex, $$\begin{align*} x^2+1 &= 0 \\ x^2&=-1 \\ x&=\pm \sqrt{-1}\\ x&= \pm i \\ \end{align*}$$
So this mysterious $\sqrt(-1)$ is at the root of our problem. Notice that we are free to call it a distance of $1$ from $0$, as well - some power of it is equal to 1, namely $i^4=1$, just like some power of $-1$ is equal to $1$, namely $(-1)^2=1$. 
Long story short, this is another case of complex numbers working entirely behind the curtain in the context of solving mysteries with why some real functions behave as they do. 
