There are three cube roots of every number [except 0, which only has 0 as all of its roots of any kind.]
i written as a complex number of the form a+bi is 0+1i.
The complex number a+bi is the vector that goes from the
origin to the point (a,b).
Therefore the number 0+1i is the vector that goes from the
origin to the point (0,1)
We want to put that number in trigonometric form.
First we draw the vector 0+1i which goes from the origin to
the point (0,1)
and we see that its angle with the right hand side of the
x-axis is 90°:
So we see that the vector's magnitude (length) is r=1, and
its angle is
Therefore the trigonometric form of i, or 0+1i, is
or in this case
Now the cosine and sine will not change if we add or subtract 360°
any whole number of times. So we can write that as:
Next we use DeMoivre's theorem, which states that
with
,
,
, since the cube root
is the
power.
Now we substitute any three consecutive values of integer n
We use the easiest ones, 0, 1 and 2
Substituting n=0
Substituting n=1
Substituting n=2
So the three cube roots of i are
,
,
Edwin