Question

Find the fourth roots of 256(cos 280° + i sin 280°).

Answer 1

Use the formula (I forgot what's it called, but):\[(\cos(a) + i\sin(a))^{x} = e^{iax}\]

Answer 2

Where does the 256 come in?

Answer 3

EULER's formula !

Answer 4

But that's an extended Euler's Formula :-)

Answer 5

e^i(280) ??

Answer 6

extended ?

Answer 7

Yes!

Answer 8

yep, e^i(280) seems fine

What you now have is

256*(e^(i*280))

You have to find 4th root of this.

Answer 9

yup.\[\sqrt[4]{ab} =\sqrt[4]{a} \cdot \sqrt[4]{b}\]

Answer 10

Ahhh! I'm so confused. How do I find the fourth root of 256*(e^(i*280))?

Answer 11

What's the non-extended formula ?

Answer 12

\[e^{i\phi} = \cos(\phi) + i\sin(\phi)\]

Answer 13

ohh, I never noticed you had put an x there. Sorry about that.

Answer 14

@shubhamsrg @ParthKohli I really need help. I still do not know how to find the fourth roots of 256(cos 280 + i sin 280)

Answer 15

I know I have 256*(e^(i*280)), but how do I find the fourth root of that?

Answer 16

\[\sqrt[4]{256(\cos280 + i\sin280)} = \sqrt[4]{256}\times \sqrt[4]{\cos(280) + i\sin(280)}\]Remember that \(\rm \sqrt[4]{stuff} = stuff^{1/4}\)?