How do you simplify #i^48 + i^150 - i^78 - i^109 + i^61#?

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Answer 1 · 2015-12-23T19:30:07

#1#. Is that a sentence?

Since #i = sqrt(-1), i^2=-1,# so #i^4=(i^2)^2=(-1)^2=1.#

Look for multiples of 4 in the exponent:
#48 = 4*12#, so #i^48=(i^4)^12=1^12=1#

#150=37*4+2,#so #i^150=(i^4)^37*i^2=1^37*(-1)=-1#
Similarly #i^78=-1.#

#109=4*54+1#, so #i^109=1^54*i=i#
Similarly #i^61=i.#

So our answer is
#i^48+i^150-i^78-i^109+i^61=#

#1+(-1)-(-1)-i+i=1.#

/ dansmath strikes again! \