Question

Which expression is a cube root of -1+i√3?

Answer 1

$$
\Large { a + bi = r (\cos \theta + i \sin \theta)
\\ where~ ~ r = \sqrt{a^2+b^2}, \theta = \arctan(b/a)

}
$$

Answer 2

where do I plug -1+i√3 in?

Answer 3

here a = -1, b = sqrt(3)

Answer 4

isn't theta=arctan(-√3/1)

-pi/3

Answer 5

we want to take the cube root of this

Answer 6

lets use demoivre's theorem

Answer 7

$$
\Large { a + bi = r (\cos \theta + i \sin \theta)
\\ where~ ~ r = \sqrt{a^2+b^2}, \theta = \arctan(b/a)
\\ \therefore \\
Given ~~~a+ bi = -1 + \sqrt3 ~i
\\r = \sqrt{(-1)^2+\sqrt3^2} = \sqrt{ 1 + 3 }=\sqrt4 = 2\\
\theta = arctan ( \frac{-\sqrt{3} } { 1}) = 120^o
\\ \therefore \\
Demoivers theorem\\
[r (\cos \theta + i \sin \theta)]^{1/n} = r^{1/n} [\cos (\theta /n) + i ~sin ( \theta /n ) ]


}
$$

Answer 8

$$
\Large {
-1 + i\sqrt3 = 2( cos ( 120^o ) + i~ sin(120^o))
\\ \therefore\\
(-1 + i\sqrt3)^{1/3} \\
=[ 2( cos ( 120^o ) + i ~sin(120^o))]^{1/3}
\\
= 2^{1/3} ( cos ( 120^o/3 ) + i ~sin(120^o/3))

}
$$

Answer 9

120/3 =40