i^-15 in standard form

Question

p0sitr0n · Answer

i^-1?

anonymous · Answer

probably easiest to understand if you know $i^{15}$ first 
do you know that one?

jonnyvonny · Answer

$$i^1=\sqrt{-1}\left|  \right| i^2=\sqrt{-1}*\sqrt{-1}=1\left|  \right| i^3=\sqrt{-1}*\sqrt{-1}*\sqrt{-1}=\sqrt{-1}$$ The pattern continues. All negative does is makes "-1" become the denominator, which, essentially doesn't change anything, because 1/1, when you take the reciprocal, is still 1/1. Anyways, every pair you can cancel out because it results in 1, which when multiplied by anything, is still the multiplied value: 1*2=2....1*10x=10x ect. 
So yes:$$i^-=i^1=i$$

anonymous · Answer

hmm i would do this
if i want $i^{15}$ i think "4 goes in to 15 with a remainder of 3 and so $i^{15}=i^3=-i$

anonymous · Answer

by the way the pattern is not quite what was written above
it is 
$$i^1=i\
i^2=-1\
i^3=-i\
i^4=1$$ and it repeats every 4

anonymous · Answer

this tells you 
$$i^{-15}=\frac{1}{-i}$$ then 
$$\frac{1}{-i}\times \frac{i}{i}=i$$