Question 172894
Take note that 49 = 48 + 1 and 48 = 12*4. 


Also, we'll use the identities {{{x^(y+z)=x^y*x^z}}} and {{{i^4=1}}}



So {{{i^(49)=i^(48+1)=i^(12*4+1)=i^(12*4)i^1=(i^4)^12*i=(1)^12*i=(1)*i=i}}}



This means that {{{i^49=i}}}


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If you want an easier method, do the following:



Divide the exponent 49 by 4 to get 12 remainder 1



Since the remainder is 1, this tells us that the answer is "i" (ie {{{i^49=i}}})




This is a handy thing to remember (or put in your reference book)


If the remainder is 0, the answer is 1
If the remainder is 1, the answer is "i"
If the remainder is 2, the answer is -1
If the remainder is 3, the answer is "-i"