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! \

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