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Topic: Root of iota
Replies: 3   Last Post: Aug 4, 1997 8:47 AM

 Messages: [ Previous | Next ]
 DrkWiz9999 Posts: 30 Registered: 12/12/04
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

Date Subject Author
8/3/97 rscas@giasdl01.vsnl.net.in
8/3/97 David Ullrich
8/4/97 DrkWiz9999
8/4/97 nobody@nowhere.on.ca