SOLUTION: {{{i^1= i}}} {{{ i^5 = i}}} {{{i^2= -1 }}} {{{ i^6 = -1}}} {{{i^3= -i }}} {{{ i^7= -i}}} {{{i^4= 1 }}} {{{ i^8 = 1}}} I am supposed to use the above pa

Algebra ->  Rational-functions -> SOLUTION: {{{i^1= i}}} {{{ i^5 = i}}} {{{i^2= -1 }}} {{{ i^6 = -1}}} {{{i^3= -i }}} {{{ i^7= -i}}} {{{i^4= 1 }}} {{{ i^8 = 1}}} I am supposed to use the above pa      Log On


   



Question 79349: i%5E1=+i +i%5E5+=+i
i%5E2=+-1+ +i%5E6+=+-1
i%5E3=+-i+ ++i%5E7=+-i
i%5E4=+1+ +i%5E8+=+1


I am supposed to use the above pattern to simplify the powers of i.
i%5E65
Please help
Please help

Found 2 solutions by chitra, josmiceli:
Answer by chitra(359) About Me  (Show Source):
You can put this solution on YOUR website!
Here we are given the values of i till i%5E8 = 1

Now to find i%5E65 this can be written as:

This can be written as:

%28i%5E13%29%5E5


Lets now find the value of i%5E13 from the given data:

i%5E8 = 1

i%5E9 = i%5E8+%2A+i

= 1+%2A+i

= i

i%5E10 = i%5E9+%2Ai

= i * i

= i%5E2

= -1

Similarly the rest are also found out.

i%5E11 = -i


i%5E12 = 1

i%5E13 = i

Thus the value of i%5E13 = i

And we know that i%5E5 = i


Thus, i%5E65 = %28i%5E13%29%5E5

= %28i%29%5E5

= i

Hence, the solution.


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2nd method


Or we can also find the value of i%5E13 = %28%28i%5E6%29%5E2%29+%2Ai

= %28%28-1%29%5E2%29%2Ai

= 1+%2A+i

= i

Now 65 = 13 * 5

i%5E65 = %28i%5E13%29%5E5

= %28i%29%5E5

= i




Hence, the solution.

Regards
Chitra
Online MAth tutor
www.knowledgeonlineservices.com



Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
i%5E1=+i +i%5E5+=+i
i%5E2=+-1+ +i%5E6+=+-1
i%5E3=+-i+ ++i%5E7=+-i
i%5E4=+1+ +i%5E8+=+1
The easiest property to make use of is
i%5E4=+1+ because every time you
multiply it times itself, you still get 1
i%5E4+%2A+i%5E4+=+1
i%5E4+%2A+i%5E4+%2A+i%5E4+=+1
i%5E4+%2A+i%5E4+%2A+i%5E4+%2A+i%5E4+=+1
Remember you ADD the exponents in each case, so
i%5E8+=+1
i%5E12+=+1
1%5E16+=+1
Think of i%5E65 as i%5E4 multiplied times itself a certain
number of times and multiply that by whatever factor is left over
65%2F4+=+16 and 1 remainder
What you've got is %28i%5E4%29%5E16+%2A+i or
1%5E16+%2A+i+=+i
So, i%5E65+=+i answer