Question

Please help. Find the cube root of -27 that graphs in the first quadrant.

3(cos?+isin?) Use degree measure.

Answer 1

`z = r (cos \theta + i sin \theta) \\ \\ \sqrt[3]{z} = \sqrt[3]{r}(cos (\theta +360k)/(3) +i sin (\theta +360k)/(3) ) , k = 0,1,2`

Now find r and theta for -27

`-27 = 27(-1 +0i) = 27(cos 180 + i sin 180)`

r = 27 , theta = 180


`\sqrt[3]{-27} = \sqrt[3]{27} (cos (180 +360k)/(3) +i sin (180 +360k)/(3) )`

`= 3(cos 60 + i sin 60) , k =0 \\ \\ = 3 (cos 180 + i sin 180) , k =1 \\ \\ =3(cos 300+i sin 300), k =2`