What is the #rootii#?
2 Answers
Oct 14, 2017
Using a similar question:
Using Euler's identity:
We know that:
Therefore,
This can be proved by calculating
Oct 14, 2017
Explanation:
Note that:
#i = e^((pi/2+2npi)i)" "# for any#n in ZZ#
So:
#root(i)(i) = i^(1/i) = i^(-i) = (e^((pi/2+2npi)i))^(-i) = e^((pi/2+2npi)*(-i^2))= e^(pi/2+2npi)#
with principal value