# Complex numbers and beyond...

1. Aug 18, 2015

### Bruno Tolentino

If the solution of the quadratic equation $$\frac{-b \pm \sqrt{b^2-4ac}}{2a}$$ produces a new kind of number, the complex numbers $$a \pm i b$$ so, the solution the cubic equation should to produce a new kind of number too, and the solution of the quartic equation too, etc...

2. Aug 18, 2015

### SteamKing

Staff Emeritus

3. Aug 19, 2015

### Raffaele

I totally agree. By the fundamental theorem of algebra an equation
$$P_n(z)=a_0+a_1z+\ldots+a_nz^n=0$$
has at least a solution in the complex field $\mathbb{C}$
$$z_1=a+bi$$
So the LHS of the equation above can be written as
$$(z-z_1)P_{n-1}(z)=0$$
we can repeat the previous step up to n times getting n roots $z_1,\;z_2,\;,\;\ldots,\;z_n$ such that
$$(z-z_1)(z-z_2)\ldots(z-z_n)=0$$

Thus the equation of degree n has up to n complex roots. They are not necessarily all distinct. Some of them can be multiple as in the following example
$$x^4(x-1)^3(x-2)^5=0$$
has the solutions
$x=0,\;4ple,\;x=1,\;3ple,\;x=2,\;5ple$

4. Aug 19, 2015

### micromass

Staff Emeritus
And that's the beauty of the complex numbers! That you don't need any new kind of number for solving higher order equations! This point of view is made very clear in the Feynman lectures on complex numbers. Because naively, you do indeed expect that for solving new kind of equations, you will need new kind of numbers. But the entire mystery and beauty of complex numbers is that they are enough to solve all polynomial equations. This turns out to be very very difficult to prove though, it is called "the fundamental theorem of algebra" and you'll need some analysis to show this.