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i^61 ???

Guest Nov 23, 2015

Best Answer 

 #1
avatar+91051 
+10

i^61

 

\(i^{61} = i^{60}\times i\\ = (i^{4})^{15} \times i\\ = [(i^2)^2]^{15} \times i\\ = [(-1)^2]^{15} \times i\\ = [1]^{15} \times i\\ = 1\times i\\ =i\)

Melody  Nov 23, 2015
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2+0 Answers

 #1
avatar+91051 
+10
Best Answer

i^61

 

\(i^{61} = i^{60}\times i\\ = (i^{4})^{15} \times i\\ = [(i^2)^2]^{15} \times i\\ = [(-1)^2]^{15} \times i\\ = [1]^{15} \times i\\ = 1\times i\\ =i\)

Melody  Nov 23, 2015
 #2
avatar+78754 
+10

Thanks, Melody for that answer....

 

Here's another way to evalute this for    i    raised to any  "n"   positve integer

 

n mod 4  =  1    →   i

 

n mod 4  =  2   →   -1

 

n mod 4  = 3   →   - i

 

n mod 4  = 4   →    1

 

So

 

i^61  =      61 mod 4   = 1    →   i            just as Melody found   !!!

 

 

 

cool cool cool

CPhill  Nov 23, 2015
edited by CPhill  Nov 23, 2015

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