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Re: Root of iota
Posted:
Aug 4, 1997 2:36 AM


>Good Day Everyone >I happen to a student of 11 th grade doing math .We have just studied >complex numbers and I wonder if anybody could help me sove this . If iota >is the root of 1 and omega is the cube root of 1 how can one derive the >square and cube roots of iota and omega > >Thank You Very much > >S.S.Rajen
Okay, here we go:
cube root of iota, sqr(omega)  if a=(1)^1/3 then sqr(a) = 1^1/6 = (1^1/3)^1/2 The third root of 1 is 1 because 1*1*1 = 1 so sqr(1) = i therefor (1)^1/6 = i
cube root of omega  if a=(1)^1/3 then a^1/3 = (1)^1/9 The third root of 1 is 1 so (1)^1/9 = (1) ^ 1/3 (1)^1/3 is 1 so (1)^1/9 is 1
sqr(iota)  sqr(1)=i so (1)^1/4 = sqr i This requires some trigonometry and calculus i = cos pi/2 + i * sin pi/2 = cis pi/2 (c * cis u )^1/n = c^1/n * cis (c/n) sqr ( cis pi/2) = cis (pi / 4) sqr (i) = cos (pi/4) + i * sin (pi/4) sqr (i) = sqr 2 /2 + i * sqr 2 /2
Now to summarize
(1)^1/4 = sqr 2 /2 + i * sqr 2/2 (1)^1/6 = i (1)^1/9 = 1
Matthew A. Halverson



