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Researching transcendental numbers I have only come across ones with a transcendental real part. I can't think of any which are pure imaginary or are not based on a real transendental number, t, of the form $t + ni$. Any ideas?

Edit: After some excellent responses, I now have another thought see Complex transcendentals not known in component form?

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Nonsematter
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    Ok, I have found one to add to the collection: $\sqrt{i\pi+\sqrt{i\pi+\sqrt{i\pi+\sqrt\cdots}}}$ thanks to zeraoulia rafik – Nonsematter Jul 21 '16 at 13:27
  • @lhf True, should've been more explicit: I'm looking for ones not based off real transcendental numbers – Nonsematter Jul 21 '16 at 13:29
  • From Gelfond-Schneider, if $\alpha \ne 0,\ 1$ and $\beta$ are algebraic numbers with $\beta$ irrational, then $\alpha^{\beta}$ is transcendantal. You can construct many complex examples from this. – Joel Cohen Jul 21 '16 at 13:30
  • @JoelCohen Nice, thanks for sharing that! – Nonsematter Jul 21 '16 at 13:31
  • See also http://math.stackexchange.com/questions/281480/is-a-complex-number-with-transcendental-imaginary-part-transcendental. – lhf Jul 21 '16 at 13:33
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    Note that if $a+bi$ is transcendantal, then at least one number among $a$ and $b$ is transcendantal. – Joel Cohen Jul 21 '16 at 13:35
  • Regarding your edit, ask a separate question. – lhf Jul 21 '16 at 13:43

4 Answers4

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We have $a + bi$ is algebraic iff $a$ and $b$ are algebraic.

Therefore, if $a + bi$ is transcendental then at least one of $a$ or $b$ is transcendental.

So, all complex transcendental numbers are "based" on real transcendental numbers.

lhf
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  • Ok, I see what you mean. So for any transcendental number, $t$, $Im(t)$ or $Re(t)$ (or both) must be transcendental. – Nonsematter Jul 21 '16 at 13:38
  • @Nonsematter, yes. – lhf Jul 21 '16 at 13:39
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    Just a comment. It's still conceivable that there exists a complex number $a + bi$ for which we can prove $a + bi$ is transcendetal, but we don't know which of $a$ and $b$ is transcendental. I think that would be very interesting. – Caleb Stanford Jul 21 '16 at 18:40
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    @6005: I guess $z=\pi e^2+\pi ei$ might be an example of this. It is transcendental because $\frac{\mathrm{Re}(z)}{\mathrm{Im}(z)}=e$ is transcendental, but it is not known whether $\pi e$ is irrational, and I guess the same for $\pi e^2$. – Jonas Meyer Jul 21 '16 at 23:09
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    @JonasMeyer: Instead of guessing about $\pi e^2$ we can use $\pi+e+\pi e i$, since both $\pi+e$ and $\pi e$ are definitely of unknown status. But at least one of them is transcendental, because otherwise the roots of $x^2-(\pi+e)x+\pi e$ would be algebraic. – hmakholm left over Monica Jul 21 '16 at 23:56
  • Thanks Jonas and Henning, those are both nice examples. – Caleb Stanford Jul 22 '16 at 00:13
  • @HenningMakholm's observation $(x-c)(x-d)=x^2-(c+d)x+cd$ shows that whenever $c$ and $d$ are two transcendental numbers, then either their sum $c+d$ or their product $cd$ (or both) must be transcendental as well. It is trivial, of course, to find examples where only one of these two is transcendental. – Jeppe Stig Nielsen Jul 22 '16 at 08:51
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Real transcendental number: $\pi$, $e$, $\ldots$
Imaginary transcendental number: $i\pi$, $ie$, $ie^\pi$ , $i\pi^e (\text{suspected to be true but not yet proved})$ $\ldots$
Complex number with transcendental real and imaginary parts: $\pi+i\pi$, $e+i\pi$, $\pi+ie$, $e+ie$, $\ldots$

Moreover, a purely imaginary transcendental number can be $\ln(-1)$, following from the equation $e^{i\pi}=-1$ and in which case, we have to extend the domain of definition and hence the range of the natural logarithm.

SchrodingersCat
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    Is it known that $\pi^e$ is transcendental? – Jonas Meyer Jul 21 '16 at 14:48
  • @JonasMeyer It is not proven but generally mathematicians believe it to be true. You can check this link ... http://sprott.physics.wisc.edu/pickover/trans.html – SchrodingersCat Jul 21 '16 at 14:53
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    Transcendental would be the safest bet, but is there a particular reason to believe that particular number is transcendental? That blog post doesn't provide any evidence. As Gerry Myerson once said, "every number not known to be algebraic is suspected transcendental." – Jonas Meyer Jul 21 '16 at 14:56
  • @JonasMeyer I totally agree with what you say. Accordingly, I have edited my answer because the expression looks too beautiful not to be a transcendental number. – SchrodingersCat Jul 21 '16 at 14:59
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    @JonasMeyer I think you're conflating "reason to believe" with "mathematical proof." – djechlin Jul 22 '16 at 00:52
  • @djechlin: What is a reason to believe? – Jonas Meyer Jul 22 '16 at 04:24
  • @JonasMeyer the axioms of ZF or the integers seem consistent, which i believe since we've been doing a lot of math without any apparent problems. – djechlin Jul 22 '16 at 06:14
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I've always been intrigued by $$i^i=e^{-π/2}$$ (at least its principal value) and you will note that $$i^{i^i}=e^{(iπ/2) e^{-π/2}}$$ does not fall into the facile category of $a+ib$ where $a$ and $b$ are well-known transcendental numbers.

Scott Centoni
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$i^{i+n}$ is transcendental for integer $n$.

Descartes Before the Horse
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