# Strange real numbers requiring use of complex numbers to exist

1. Jun 11, 2012

### daniel.e2718

I couldn't really think of a good title for this question, lol.

Is it possible that a real number exists that can only be expressed in exact form when that form must includes complex numbers?

For example, the equation

$2 \, x^{3} - 6 \, x^{2} + 2 = 0$

has the following roots

$x_1 = -\frac{1}{2} \, {\left(\frac{1}{2} i \, \sqrt{3} + \frac{1}{2}\right)}^{\left(\frac{1}{3}\right)} {\left(i \, \sqrt{3} + 1\right)} - \frac{-i \, \sqrt{3} + 1}{2 \, {\left(\frac{1}{2} i \, \sqrt{3} + \frac{1}{2}\right)}^{\left(\frac{1}{3}\right)}} + 1$

$x_2 = -\frac{1}{2} \, {\left(-i \, \sqrt{3} + 1\right)} {\left(\frac{1}{2} i \, \sqrt{3} + \frac{1}{2}\right)}^{\left(\frac{1}{3}\right)} - \frac{i \, \sqrt{3} + 1}{2 \, {\left(\frac{1}{2} i \, \sqrt{3} + \frac{1}{2}\right)}^{\left(\frac{1}{3}\right)}} + 1$

$x_3 = {\left(\frac{1}{2} i \, \sqrt{3} + \frac{1}{2}\right)}^{\left(\frac{1}{3}\right)} + \frac{1}{{\left(\frac{1}{2} i \, \sqrt{3} + \frac{1}{2}\right)}^{\frac{1}{3}}} + 1$

which have numerical approximations of

$x_1 \approx 0.65270364$

$x_2 \approx -0.53208889$

$x_3 \approx 2.8793852$

When I run these through Sage's simplify_full() command (five times, even), they just become single fractions.

Is this a CAS simplification thing or are there real numbers that simply cannot exist without complex numbers?

2. Jun 11, 2012

But you just expressed those numbers in a form without complex numbers.

Anyway, I don't think such numbers exist, since any complex algebra can be replaced with geometrical operations in polar space, involving just real numbers.

3. Jun 11, 2012

### Number Nine

It's entirely possible for a polynomial with coefficients in the reals to have one or more complex roots, but this probably isn't what you're asking. The simple answer is that it's not necessary to reference the complex numbers in order to describe or obtain the reals; the complex numbers are what is known as an extension of the real number field, created for reasons that have to do largely with polynomials (at least, as far as mathematicians are concerned).

4. Jun 12, 2012

### haruspex

The interesting question was the one at the start of the OP:
That is, can there be an expression using rationals, surds and complex numbers that evaluates to a real but cannot be expressed using only real rationals and surds?
I suspect the answer is yes. You could take any such complex expression, whether or not it evaluates to a real, and add its complex conjugate (by changing i to -i everywhere) to produce one that does evaluate to a real.
I took the specific case here of (1+i√3) and wrote it as (a+b√3+ic+id√3)3. It looked like it might be possible to show that a, b, c and d cannot all be rational.

5. Jun 12, 2012

### JJacquelin

Hello daniel.e2718 !

The roots written without any complex term :
x1 = 1-2cos(4pi/9)
x2 = 1-2cos(2pi/9)
x3 = 1-2cos(pi/9)

6. Jun 12, 2012

### Hurkyl

Staff Emeritus
7. Jun 12, 2012

### HallsofIvy

Just to be clear, real numbers "exist" irrespective of any way of writing them.

However, it is true that there exist cubic equations, having real solutions, such that if you use Cardano's cubic formula, which involves taking square roots and then cube roots, you can wind up with complex numbers in the intermediate calculations- the imaginary parts then eventually cancelling out. It was this discovery that led to complex numbers being accepted as "numbers".
(I almost wrote "as real numbers"!)

8. Jun 12, 2012

### daniel.e2718

Ahh this. I didn't think about that at all. Then again, de Moivre's formula...

But also, cosine is a transcendental function. None of the exact numbers in my original post had infinite series.

Is it possible to express the above numbers in an exact form without using transcendental functions or using an imaginary unit? Or transcendental numbers like pi.

This is interesting