# How do you simplify #i^38#?

##### 1 Answer

Feb 10, 2016

#### Explanation:

Let's see what happens if we compute any power of

#i = i#

#i^2 = -1#

#i^3 = i^2 * i = -1 * i = -i#

#i^4 = i^2 * i^2 = -1 * (-1) = 1#

#i^5 = i^4 * i = 1 * i = i#

... and so on, after that, the sequence

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How can you determine which one it is for

Let's start with factorizing

#i^38 = i^(4*9+2) = i^(4*9) * i^2 = (i^4)^9 * i^2 = 1^9 * (-1) = -1#

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*Let me also additionally show you a general way to determine #i^n# for any positive integer #n#.*

*There are four possibilites:*

if#n# can be divided by#4# , then#i^n = 1# if#n# can be divided by#2# (but not by#4# ), then#i^n = -1# if#n# is an odd number but#n-1# can be divided by#4# , then#i^n = i# if#n# is an odd number but#n+1# can be divided by#4# , then#i^n = -i#

*Described in a more formal way,*

#i^n = {(1, " " n= 4k),(i, " " n = 4k + 1),(-1, " " n = 4k + 2),(-i, " " n= 4k + 3) :}#

for#k in NN_0# .