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25 49. If sumi^((2n+1)!) = a + ib, where...

25 49. If `sumi^((2n+1)!) = a + ib`, where `i = sqrt-1`, then `a-b`, is

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Sum of four consecutive powers of i(iota) is zero. i.e., i^(n)+i^(n+1)+i^(n+2)+i^(n+3)=0,forall n in I. If sum_(n=1)^(25)i^(n!)=a+ib, " where " i=sqrt(-1) , then a-b, is

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