# How do you simplify #i^15#?

##### 1 Answer

#### Explanation:

Remember that

Thus,

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

Also, remember the power rule

#a^m * a^n = a^(m+n)#

Thus, you have

#i^15 = i^(4 + 4 + 4 + 3) = i^4 * i^4 * i^4 * i^3 = 1 * 1 * 1 * i^3 = i^3 = i^2 * i = -1 * i = -i#

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Also, I'd like to offer you a more general solution for

Try to recognize the pattern:

#i = i#

#i^2 = -1#

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

#i^4 = i^3 * i = -i * i = -i^2 = 1#

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

#i^6 = i^4 * i^2 = -1#

...

So, basically, the power of

Thus, to compute

- 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# .