complex numbers
I have recently learnt about complex numbers, but there still has been a question in my head:
Is there a number $k$, where $k$ is in the form of $bi$ (where $b$ is an integer), such that when $n$ is a positive integer and $r$ is a negative integer, we have
$$n^k=r?$$
Note: I've changed the question, so some comments may not make sense.
[begin crackpot theory]
Mathematicians have committed a great blunder by thinking that a certain imaginary number is $i$ and another is $-i$, when it's really the other way around!!!!
[end crackpot theory]
If one were to take the position above, and try to rewrite all of mathematics consistently with the theory propounded above, the result would be that nothing at all would change. Which one is called $i$ and which is called $-i$ doesn't matter.
It does in some contexts make sense to pay attention to whether the real part of a complex number is positive. A Dirichlet series $\displaystyle\sum_{n=1}^\infty\frac{a_n}{n^s}$ converges if the real part of $s$ is greater than the real part of the abscissa of convergence of the series, and diverges if the real part of $s$ is less than the abscissa of convergence.
To Henry's comment here is the drawing:
Best Answer
The answer is no. In fact, there are no integers $n > 1, b \ne 0$ for $n^{ib}$ to be real.
If that happens,
$$\Im n^{ib} = \Im e^{ib\log n} = \sin(b\log n) = 0$$ implies we can find an integer $N \ne 0$ so that
$$\log(n)b = N\pi\quad\implies\quad(e^\pi)^N = n^b\quad\implies\quad e^\pi \text{ is algebraic}.$$
This contradicts with the fact that the Gelfond's constant $e^{\pi}$ is transcendental.