I need help finding the roots??

[TheNascar92]

I have this equation:

-4-4(root 3i) >>this reads negative four minus 4(root 3i) I am supposed the find the cube roots of this so it would look like:

x^3=-4-4(root 3i)

the answers i found were: 2(cos80+isin80), 2(cos200+isin200), 2(cos320+isin320)

Im not sure if these are correct? And how would i write them in standard form (a+bi form) if the cosine/sine values are not exact? could i just use a calculator to get a decimal? thank you!

 

[soroban]

Helolo, TheNascar92!

\(\displaystyle \text{Find the cube roots of: }\;-4 -4\sqrt{3}\,i\)

\(\displaystyle \text{Write answers in the form: }\:a + bi\)

\(\displaystyle \text{The answers i found were: }2(\cos80^o+i\sin80^o),\; 2(\cos200^o+ i\sin200^o),\; 2(\cos320^o+ i\sin320^o)\)
These are correct!

\(\displaystyle \text{How would i write them in standard form }a+bi\text{ form if the cosine/sine values are not exact?}\)

You already have them in standard form.
. . But radians should be used.


\(\displaystyle \text{We have: }\;-4-4\sqrt{3}\,i \;=\;8\bigg[\cos\left(\tfrac{4\pi}{3}+2\pi n\right) + i\sin\left(\tfrac{4\pi}{3}+ 2\pi n\right)\bigg]\)

\(\displaystyle \text{Then: }\;(-4-4\sqrt{3}\,i)^{\frac{1}{3}} \;=\;2\bigg[\cos\left(\tfrac{4\pi}{9} + \tfrac{2\pi}{3}n\right) + i\sin\left(\tfrac{4\pi}{9} + \tfrac{2\pi}{3}n\right)\bigg]\quad \text{ for }n \:=\:0,1,2\)


\(\displaystyle \text{Therefore: }\;2\left(\cos\tfrac{4\pi}{9} + i\sin\tfrac{4\pi}{9}\right),\;\;2\left(\cos\tfrac{10\pi}{9} + i\sin\tfrac{10\pi}{9}\right),\;\;2\left(\cos\tfrac{16\pi}{9} + i\sin\tfrac{16\pi}{9}\right)\)

 

[TheNascar92]

Thanks a million

 

[shadowleira]

i still need an explanation on how to find cube root though