*not of unity*?

Say: (1 + i)

^{1/3}

For this example, we use:

w

Note:

The roots are then defined from:

w

w

_{n}=^{3}Ör [cos(q + 2 pk/n) + isin[(q + 2 pk/n)]Note:

^{3}Ör= r^{1/3}= [Ö2]^{1/3}= (2)^{1/ 6}The roots are then defined from:

w

_{n}= (2)^{1/ 6}[cos(q + 2 pk/3) + isin(q + 2 pk/3)]Since q = arctan y/x = 45 deg then, q = p/4

So the consecutive roots are(in order):

w

_{0}= (2)

^{1/6}[cos(p/4 + 2 pk/3) + isin(p/4 + 2 pk/3)]

Since k=0:

w

_{0 }= (2)

^{1/6}[cos(p/4) + isin(p/4)]

Further:

w

Working on the interior brackets (adding fractions in p):

(p/4 + 2 p/3) = (3 p + 8p)/ 12 = 11 p/12

So:

w

and the answer can be left in this form, OR the reader can obtain cos(11p/12) and sin(11p/12) and work out the specific numbers.

_{1}= (2)^{1/6}[cos(p/4 + 2 pk/3) + isin(p/4 + 2 pk/3)]Working on the interior brackets (adding fractions in p):

(p/4 + 2 p/3) = (3 p + 8p)/ 12 = 11 p/12

So:

w

_{1 }= (2)^{1/6}[cos(11 p / 12) + isin(11 p/12)]and the answer can be left in this form, OR the reader can obtain cos(11p/12) and sin(11p/12) and work out the specific numbers.

Finally since k =2 (for w

w

Again, adding the fractions in p in the interior brackets:

(p/4 + 4p/3) = (3 p + 16 p)/ 12 = 19 p/12

_{2}):w

_{2}= (2)^{1/6}[cos(p/4 + 4p/3) + isin(p/4 + 4p/3)]Again, adding the fractions in p in the interior brackets:

(p/4 + 4p/3) = (3 p + 16 p)/ 12 = 19 p/12

So:

w

_{2 }= (2)^{1/6}[cos(19 p/12 ) + isin(19 p/12)]Problems for the Math Maven:

1) Find all the roots of: (-2 + 2i)

^{1/3}
2) Find all the roots of (-16)

^{1/4}
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