11.4. De Moivre’s Theorem and nth Roots http://www.ck12.org
Guided Practice
- Check the three cube roots of 8 to make sure they are truly cube roots.
- Solve forzby finding the nth root of the complex number.
z^3 = 64 − 64
√
3 i
- Use De Moivre’s Theorem to evaluate the following power.
(√
2 −√ 2 i
) 6
Answers:
z^31 = 23 = 8
z^32 =
(
− 1 +i√ 3
) 3
=
(
− 1 +i
√
3
)
·
(
− 1 +i
√
3
)
·(− 1 +i
√
3 )
=
(
1 − 2 i√ 3 − 3
)
·
(
− 1 +i√ 3
)
=
(
− 2 − 2 i
√
3
)
·
(
− 1 +i
√
3
)
= 2 − 2 i
√
3 + 2 i
√
3 + 6
= 8
Note how many steps and opportunities there are for making a mistake when multiplying multiple terms in
rectangular form. When you checkz 3 , use trigonometric polar form.
z^33 = 23 cis
(
3 ·^43 π
)
= 8 (cos 4π+i·sin 4π)
= 8 ( 1 + 0 )
= 8
- First write the complex number in cis form. Remember to identifyk= 0 , 1 ,2. This means the roots will appear
every^3603 ◦= 120 ◦.
z^3 = 64 − 64
√
3 i= 128 ·cis 300◦
z 1 = 12813 ·cis
( 300
3 ◦
)
= 12813 ·cis( 100 ◦)
z 2 = 12813 ·cis( 220 ◦)
z 3 = 12813 ·cis( 340 ◦)
- First write the number in trigonometric polar form, then apply De Moivre’s Theorem and simplify.