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Someone recently told me certain real numbers could only be expressed in closed form via an expression involving complex numbers.

Is this true? If so do these numbers have a name?

What is a simple example?

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    $i^i$, where $i$ is the imaginary unit, comes to mind. It is a real number, with value $e^{-\pi/2}$, but it doesn't fit your example because I just also gave a closed form with no imaginary units. Interesting. – 2017-02-22
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    @OmnipotentEntity: I would like you to ask to the person who told you this, to give at least one such example. Is it possible ? – 2017-02-22
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    He didn't know of an example. He said he had read it somewhere. – 2017-02-22
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    Note that this needs a very specific definition of 'closed-form', but it's true under that definition; 'nested radicals of rational expressions' is a pretty good version. – 2017-02-22
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    That seems to be that that doesn't make any sense. I could be wrong but if it's expressible in closed for in terms of non-real complex numbers and a nonreal complex number is expressible in terms of reals and square roots of negative reals, then all reals and indeed all complex numbers that are expressible in any closed form are expressible in closed form via reals. – 2017-02-22
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    There is no such thing as "expressed via". Once you try to make it precise, it disappears. Other than that, the accepted answer is a nice example. – 2017-02-22

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This phenomenon occurs prominently in the (historically interesting) casus irreducibilis of third degree polynomial equations with three real roots. Take the equation $$x^3-2x^2-6x+5=0$$ as an example.

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To answer my final part about a nice, minimal example:

$$\sqrt[3]{1+i \sqrt{7}}+\sqrt[3]{1-i \sqrt{7}}$$

Which $\approx$ 2.6016791318831542525

This value was adapted from one of the roots of the polynomial posted in the accepted answer, then back solved for the generating cubic to ensure it is a real casus irreducibilis. (Previous version using $\sqrt{5}$ did not have a rational cubic.)

The generating cubic is simply $x^3 - 6x - 2 = 0$.

As noted by a @Steven Stadnicki , this is only irreducible under the radicals. Using Euler's formula we can find a trigonometric representation of this number.

$$2 \sqrt{2} \cos{\left(\frac{\tan^{-1}{\left(\sqrt{7}\right)}}{3}\right)}$$

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$2$ things to be noted:

  • The set of real numbers $(\mathbb{R})$ is a subset of the set of complex numbers $(\mathbb{C})$. Any real number is a complex number but the converse is not true.

  • If any number (real or not) can be expressed involving certain non-zero complex expressions, then it is a complex number, by definition.

If this is not what you intended to ask, then please tell me.

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    I think that for the first point, the word "always" should be removed. – 2017-02-22
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    @Adren Why do you think so? I think there is some kind of misinterpretation from my side. Isn't the converse true in some cases? – 2017-02-22
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    This is probably a personal failing, but I am unable to see how this answers any aspect of my question. Perhaps I failed to make myself clear. The question is perhaps better stated as "Is there a real number (ie, a complex number with imaginary part = 0), which can only be expressed in terms of a combination of complex terms whose imaginary parts cannot be eliminated through any method of simplification?" For instance the number $i^i$ is a real number expressed in imaginary units, but it can be simplified to eliminate the complex part. Is there a similar number which cannot? – 2017-02-22
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    @SchrodingersCat: I believe that we agree fundamentally ... but the natural language is sometimes ambiguous. What is the converse of "any real number is a complex number" ? The sentence "any complex number is a real number" is false and the sentence "some complex numbers are real" is false ... I don't see any reason to add "always". But maybe I miss something : you should know that my language is not english, but french :) – 2017-02-22
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    @Adren, if $(\mathbb{R})$ is a subset of $(\mathbb{C})$, then some complex numbers are real. They're just better described by the term "real numbers" than "complex numbers with no imaginary part." – 2017-02-22
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    @OmnipotentEntity Maybe you would like to explain what do you mean by " a real number expressed in imaginary units but it can be simplified to eliminate the complex part"? – 2017-02-22
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    @OmnipotentEntity If its a real number then how is it expressed in terms of imaginary unit? Isn't it against the very definition of a real number and complex number? – 2017-02-22
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    @SchrodingersCat FWIW, the question is perfectly clear to me. It might be better expressed as 'there are real numbers with a closed-form expression in terms of radicals of ('rational') complex numbers which have no closed-form expression involving only radicals of (rational) real numbers.' – 2017-02-22
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    @SchrodingersCat For instance, consider $r=\sqrt[3]{1+i}+\sqrt[3]{1-i}$. This $r$ is a real number (can you see why?) but it doesn't have an immediately obvious expression in terms of only radicals of (rational) other real numbers. (In this case, $r$ actually does have such an expression, but it's far from obvious; in other cases, as the other answer suggests, there are actually no such expressions.) – 2017-02-22
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    I just discover the typo in my previous comment !! Of course I meant : the sentence "some complex numbers are real" is TRUE ... – 2017-02-22
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    You raise a good point and we should instruct a poster when one's question is poorly formed. But I don't think we do any good to take on the french waiter attitude of pretending we have no idea what a poster is asking because his mastery of language is not as good as ours. "If this is not what you intended to ask, then please tell me." It's seems to me the poster meant, there are real numbers only expressible as expressions involving non-real complex numbers. – 2017-02-22
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    @StevenStadnicki the r you are talking about is out and out a real number just as you have said. I don't know if the question is asking for such r since this representation of r as a sum of two complex cube roots does not imply from any angle that it is a complex number with non zero imaginary part. The $i^i$ example is nice, and I understand what the question demands but the example you gave is not relevant. – 2017-02-22
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    @SchrodingersCat I disagree, it is very relevant - it is an example of a real number with a closed form expression that (i) uses non-real complex numbers and (ii) does not have an obvious conversion to an expression *not* using non-real complex numbers. It's not quite an answer to the question, since it does have a closed form expression not using non-real complex numbers, but it is definitely positive *evidence* towards an affirmative answer to the question. – 2017-02-22