197k views
5 votes
Which expression is a cube root of 2?

A) Cube root of 2 (cos (120 degrees) + i sin (120 degrees))
B) Cube root of 2 (cos (40 degrees) + i sin (40 degrees))
C) Cube root of 2 (cos (280 degrees) + i sin (280 degrees)) ...?

User Ayon
by
8.3k points

2 Answers

2 votes

Answer with explanation:


z=2^{(1)/(3)}\\\\z^3=2\\\\z^3=2 * (1 + 0 i)\\\\z^3=2 * [\cos 0^(\circ) + i \sin 0^(\circ)]\\\\z^3=2 * [\cos (2 k\pi+ 0^(\circ)) + i \sin(2 k\pi+0^(\circ))],where,k=0,1,2,\\\\z=[2*[ (\cos (2 k\pi+ 0^(\circ)) + i \sin(2 k\pi+0^(\circ))]]^{(1)/(3)}\\\\z=2^{(1)/(3)}*[\cos (2 k\pi+ 0^(\circ))/(3) + i \sin((2 k\pi+0^(\circ)))/(3)]\\\\Z_(0)=2^{(1)/(3)},for, k=0\\\\z_(1)=2^{(1)/(3)}*[\cos ((2\pi))/(3) + i \sin((2\pi))/(3)]


z_(1)=2^{(1)/(3)}*[\cos(120^(\circ))+i\sin(120^(\circ))],\text{Used Demoiver's theorem}[\cos A + i\sin A]^k=\cos Ak + i\sin Ak

Option A: →Cube root of 2 [cos (120 degrees) + i sin (120 degrees)]

User Leonid Shifrin
by
8.0k points
2 votes
I'm 100% sure that this is the expression which represents a cube root of 2: A) Cube root of 2 (cos (120 degrees) + i sin (120 degrees))
User Edward Kmett
by
8.9k points