Question

HELP PLEASE!!

Find the cube roots of 8(cos 216° + i sin 216°).

Answer 1

take the cube root of 8, that is 2
then divide each angle by 3

Answer 2

that is why this is easy, why you want to write a complex number in trig form
\[216\div 3=72\] so one answer is
\[2\left(\cos(72)+i\sin(72)\right)\]

Answer 3

please you have to apply the De Moivre formula, namely:
\[\sqrt[3]{8}(\cos (\frac{ 216+2k 180 }{ 3 })+i \sin(\frac{ 216 +2k180}{ 3 }))\]
and substitute k=0,1,2
so you should get three square roots

Answer 4

to find the other one, go around the circle again
i.e. add 360 to 216, then divide the result by 3
lather
rinse repeat

Answer 5

[8^(1/3) [\cos1/3(216\deg+360k) + i \sin 1/3(216\deg+360k)]

Answer 6

Right?

Answer 7

right!

Answer 8

for example, for k=0, you should get this:
\[2(\cos(72)+i \sin(72))\]

Answer 9

So,
= 2[cos (72deg+ 120degk) + i sin (72deg + 120degk)]
Right?

Answer 10

So,
k = 0: 2(cos 72deg + i sin 72deg)
right?

Answer 11

no, please if I set k=1 in my formula, I get this:
\[2\left( \cos \left( \frac{ 216+1*360 }{ 3 } \right)+i \sin \left( \frac{ 216+1*360 }{ 3 } \right) \right)\]

Answer 12

oh ok

Answer 13

and if I set k=2, I get:
\[2\left( \cos \left( \frac{ 216+2*360 }{ 3 } \right)+ i \sin \left( \frac{ 216+2*360 }{ 3 } \right) \right)\]

Answer 14

So,
k = 2: 2(cos 312deg + i sin 312deg)

Answer 15

that's right!

Answer 16

yes!

Answer 17

yes!

Answer 18

thank you!

Answer 19

thank you!