11.4. De Moivre’s Theorem and nth Roots http://www.ck12.org
You can see at this point that to findsyou need to take thenthroot ofr. The trickier part is to find the angles,
becausen·βcould be any angle coterminal withθ. This means that there arendifferentnthroots ofz.
n·β=θ+ 2 πk
β=θ+n^2 πkThe numberkcan be all of the counting numbers including zeros up ton−1. So if you are taking the 4throot, then
k= 0 , 1 , 2 ,3.
Thus thenthroot of a complex number requiresndifferent calculations, one for each root:
v=√nr·cis(θ+n^2 πk)f or{kεI| 0 ≤k≤n− 1 }
Example A
Find the cube root of the number 8.
Solution: Most students know that 2^3 =8 and so know that 2 is the cube root of 8. However, they don’t realize
that there are two other cube roots that they are missing. Remember to write outk= 0 , 1 ,2 and use the unit circle
whenever possible to help you to find all three cube roots.
8 =8 cis 0= (s·cisβ)^3
z 1 = 2 ·cis( 0 + 2 π· 0
3)
=2 cis 0= 2 (cos 0+i·sin 0) = 2 ( 1 + 0 ) = 2z 2 = 2 ·cis( 0 + 2 π· 1
3)
=2 cis( 2 π
3)
= 2
(
cos( 2 π
3)
+i·sin( 2 π
3))
= 2
(
−^12 +
√ 3
2 i)
=− 1 +i√
3
z 3 = 2 ·cis( 0 + 2 π· 2
3)
=2 cis( 4 π
3)
= 2
(
cos( 4 π
3)
+i·sin( 4 π
3))
= 2
(
−^12 −
√
3
2 i)
=− 1 −i√
3
The cube roots of 8 are 2,− 1 +i
√
3 ,− 1 −i√
3.
Example B
Plot the roots of 8 graphically and discuss any patterns you notice.
Solution: