Which expression is a cube root of -2i?

1) 3sqrt2(cos(210))+isin(210))

2) 3sqrt2(cos(260))+isin(260))

3) 3sqrt2(cos(90))+isin(90))

4) 3sqrt2(cos(60))+isin(60))

To use De Moivre's theorem, we first write -2i is cis form: 0 - 2i has r = 2 and theta = 270.
Then we take the cube root, which means the new result will have r^(1/3), and the angle (theta/3). This means r = cbrt(2) and theta = 90.
This means that the answer is (cube root of 2)(cos90 + i*sin90), choice C.

I suspect the choices "3sqrt2" is actually a cube root of 2, not 3 multiplied by the square root of 2.

Answer Link

Аноним Аноним · Answer 2 · 2019-05-30T14:22:03+02:00

Answer with explanation:

[tex]Z=(-2 i)^{\frac{1}{3}}\\\\Z^3= -2 i\\\\Z^3=2[0- i]\\\\z^3=2[\cos(\frac{3\pi }{2})+i \sin(\frac{3\pi }{2})]\\\\z=2^{\frac{1}{3}}[\cos(\frac{3\pi }{2})+i \sin(\frac{3\pi }{2})]^{\frac{1}{3}}\\\\z=2^{\frac{1}{3}}[\cos(2k\pi +\frac{3\pi }{2})+i \sin(2k\pi +\frac{3\pi }{2})]^{\frac{1}{3}}\\\\z=2^{\frac{1}{3}}[\cos(\frac{2k\pi}{3} +\frac{\pi }{2})+i \sin(\frac{2k\pi}{3} +\frac{\pi }{2})]^{\frac{1}{3}}\\\\ \text{use Demoiver's theorem}}\\\\ (\cos A + \sin A)^n=\cos nA +i\sin nA\\\\ \text{for, k=0,}}[/tex]

we will get one root of

[tex](-2 i)^{\frac{1}{3}},\text{which is equal to }}=2^{\frac{1}{3}}[\cos(\frac{\pi }{2})+i \sin\frac{\pi }{2})][/tex]

Option 3

Which expression is a cube root of -2i? 1) 3sqrt2(cos(210))+isin(210)) 2) 3sqrt2(cos(260))+isin(260)) 3) 3sqrt2(cos(90))+isin(90)) 4) 3sqrt2(cos(60))+isin(60))

Respuesta :

Otras preguntas

Which expression is a cube root of -2i?

1) 3sqrt2(cos(210))+isin(210))

2) 3sqrt2(cos(260))+isin(260))

3) 3sqrt2(cos(90))+isin(90))

4) 3sqrt2(cos(60))+isin(60))