How do you simplify i^48 + i^150 - i^78 - i^109 + i^61i48+i150−i78−i109+i61?

1 Answer
dansmath · George C.
Dec 23, 2015

11. Is that a sentence?

Explanation:

Since i = sqrt(-1), i^2=-1,i=√−1,i2=−1, so i^4=(i^2)^2=(-1)^2=1.i4=(i2)2=(−1)2=1.

Look for multiples of 4 in the exponent:
48 = 4*1248=4⋅12, so i^48=(i^4)^12=1^12=1i48=(i4)12=112=1

150=37*4+2,150=37⋅4+2,so i^150=(i^4)^37*i^2=1^37*(-1)=-1i150=(i4)37⋅i2=137⋅(−1)=−1
Similarly i^78=-1.i78=−1.

109=4*54+1109=4⋅54+1, so i^109=1^54*i=ii109=154⋅i=i
Similarly i^61=i.i61=i.

So our answer is
i^48+i^150-i^78-i^109+i^61=i48+i150−i78−i109+i61=

1+(-1)-(-1)-i+i=1.1+(−1)−(−1)−i+i=1.

/ dansmath strikes again! \

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