Can I write $i^{\sqrt{i} }$ more simple than this?
$i^{\sqrt{i} }=(e^{i\frac{\pi}{2}})^{e^{i\frac{\pi}{4}}}=\frac{e^{i\frac{\pi}{2}\cos\frac{\pi}{4}}}{e^{\frac{\pi}{2}\sin\frac{\pi}{4}}}$
Asked
Active
Viewed 101 times
1
-
$\cos (\pi/4) = \sin (\pi/4) = 1/\sqrt {2} $. – user76284 May 28 '17 at 16:59
-
You can apply Euler's formula again with the numerator. Always try for a form $u + iv$ – Kaynex May 28 '17 at 17:03
-
$$\frac{e^{i\frac{\sqrt{2}\pi}{4}}}{e^{\frac{\sqrt{2}\pi}{4}}}$$ Not much more. – Rafa Budría May 28 '17 at 17:03
-
See https://math.stackexchange.com/questions/3315/what-is-sqrti – lab bhattacharjee May 28 '17 at 17:05
-
Hmmm... How do you define $$i^{\sqrt{i}}$$ already? Being able to put symbols side by side does not mean the result has a meaning. Even before this, what is $$\sqrt{i}\ ?$$ – Did May 28 '17 at 17:11
1 Answers
1
Possibly one is expected to find all the values, which produces an infinite list: more generally, replacing the two square roots of $i$ by a general constant $a$, the ambiguity of complex logarithms gives $$ i^a \;=\; (e^{\log i})^a \;=\; (e^{{\pi i \over 2} +2\pi in})^a \;=\; e^{{\pi i a\over 2}}\cdot e^{2\pi ina} $$ for all integers $n$. Thus, if the exponent $a$ is not a rational number, there are infinitely-many different such expressions.
(Yes, in such a context even $1^a$ could have infinitely-many different values, but that's not in conflict with the idea that $1^n=1$ for integers $n$, nor with the evaluate-by-limit evaluations of real powers of positive real numbers, which are simply different contexts.)
paul garrett
- 52,465