#2**+13 **

$$\\-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$$$

Melody Oct 3, 2014

#2**+13 **

Best Answer

$$\\-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$$$

Melody Oct 3, 2014

#3**0 **

WOW! OUR SUPERSTAR EXPLAINED THE ANSWER, AND MY ANSWER IS CORRECT TOO!!!

WOW MELODY! YOU ARE A SUPERMODEL!!!!!

DragonSlayer554 Oct 3, 2014

#7**0 **

when melody has the chance to explain an aswer, she DEEPLY EXPLAINS IT.

DragonSlayer554 Oct 3, 2014

#8

#9**0 **

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???

DragonSlayer554 Oct 3, 2014