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what does -i equals to?

 Oct 2, 2014

Best Answer 

 #2
avatar+95369 
+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?

 

$$\\i^{-4n+3}\\
=i^{-4n}*i^3\\
=\frac{1}{i^{4n}}*i^3\\
=\frac{1}{(i^{4})^n}*i^3\\
=\frac{1}{(1)^n}*i^3\\
=\frac{1}{1}*i^3\\
=i^3\\
=i^2*i\\
=-i\\\\
so\\\\
-i=i^{4n+3} \qquad $ where $ n \in Z\\
$n can be any integer, it does not have to be positive$$$

.
 Oct 3, 2014
 #1
avatar+8263 
+8

$$-i=$$   $$-{\mathtt{1}}{i}$$

My guess is $$-{\mathtt{1}}{i}$$.

 Oct 3, 2014
 #2
avatar+95369 
+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
avatar+8263 
0

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

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

 Oct 3, 2014
 #4
avatar+95369 
+3

Yes I think that we are both right.   

 Oct 3, 2014
 #5
avatar+8263 
0

YAY!

 Oct 3, 2014
 #6
avatar+94619 
+3

Very nice proof, Melody !!! 

 

 Oct 3, 2014
 #7
avatar+8263 
0

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

 Oct 3, 2014
 #8
avatar+95369 
+3

Thanks Chris and dragon.

I was just playing.  It is good to do that sometimes.  

 Oct 3, 2014
 #9
avatar+8263 
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???

 Oct 3, 2014

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