+118690 $$\\-i=-\sqrt{-1}\\
or\\
-i=-1*\sqrt{-1}\\
-i=i^2*i\\
-i=i^3\\
or\\
-i=(-1)^3*i\\
-i=(i^2)^3*i\\
-i=i^6*i\\
-i=i^7\\
$continuing with this pattern I think$\\
-i=i^{4n+3}\qquad $Where n \in Z $ and $ n\ge 0$$
Now it has got me thinking more.
what if n was a negative integer? Would that work?
$$\\i^{-4n+3}\\
=i^{-4n}*i^3\\
=\frac{1}{i^{4n}}*i^3\\
=\frac{1}{(i^{4})^n}*i^3\\
=\frac{1}{(1)^n}*i^3\\
=\frac{1}{1}*i^3\\
=i^3\\
=i^2*i\\
=-i\\\\
so\\\\
-i=i^{4n+3} \qquad $ where $ n \in Z\\
$n can be any integer, it does not have to be positive$$$
+8262 $$-i=$$ $$-{\mathtt{1}}{i}$$
My guess is $$-{\mathtt{1}}{i}$$.
.+118690 $$\\-i=-\sqrt{-1}\\
or\\
-i=-1*\sqrt{-1}\\
-i=i^2*i\\
-i=i^3\\
or\\
-i=(-1)^3*i\\
-i=(i^2)^3*i\\
-i=i^6*i\\
-i=i^7\\
$continuing with this pattern I think$\\
-i=i^{4n+3}\qquad $Where n \in Z $ and $ n\ge 0$$
Now it has got me thinking more.
what if n was a negative integer? Would that work?
$$\\i^{-4n+3}\\
=i^{-4n}*i^3\\
=\frac{1}{i^{4n}}*i^3\\
=\frac{1}{(i^{4})^n}*i^3\\
=\frac{1}{(1)^n}*i^3\\
=\frac{1}{1}*i^3\\
=i^3\\
=i^2*i\\
=-i\\\\
so\\\\
-i=i^{4n+3} \qquad $ where $ n \in Z\\
$n can be any integer, it does not have to be positive$$$
+8262 WOW! OUR SUPERSTAR EXPLAINED THE ANSWER, AND MY ANSWER IS CORRECT TOO!!!
WOW MELODY! YOU ARE A SUPERMODEL!!!!!
+8262 when melody has the chance to explain an aswer, she DEEPLY EXPLAINS IT.
+118690 Thanks Chris and dragon.
I was just playing. It is good to do that sometimes. 
+8262 playing? i dont think so. you did a great answer, so that is no joke! are you one of those super genious people on Earth???
