If `iota=sqrt(-1)` then `iota^3=?`

If |z-iota|=sqrt(2) then amp((z-1)/(z+1)) equals

Let f(x)=x^(3)+ax^(2)+bx+c be a cubic polynomial with real coefficients and all real roots.Also |f(iota|=1 where iota=sqrt(-1) statement-1: All 3roots of f(x)=0 are zero Statement-2: a+b+c=0

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_(r=-2)^(95)i^(r)+sum_(r=0)^(50)i^(r!)=a+ib, " where " i=sqrt(-1) , the unit digit of a^(2011)+b^(2012) , is

bar(a)+2bar(b)+3bar(c)=bar(0) and bar(a)xxbar(b)+bar(b)xxbar(c)+bar(c)xxbar(a)=iota(bar(b)xxbar(c)), then iota=

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

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_(r=4)^(100)i^(r!)+prod_(r=1)^(101)i^(r)=a+ib, " where " i=sqrt(-1) , then a+75b, is

sqrt(5+12 iota)

Find the value of   ( iota^2 - iota ) times ( iota^2 + iota ) times ( iota^3 + iota ) is

Determine the value of - ((1+iota)/sqrt2)^8 + ((1-iota)/sqrt2)^8 (a) 1 (b) 2 (c) 3 (d) 0

If z_(1),z_(2),z_(3) are vertices of an equilateral triangle inscribed in the circle |z|=2 and if z_(1)=1+iota sqrt(3), then