Determine-the-fourth-roots-of-16-giving-the-results-in-the-form-a-jb-

Question Number 14588 by tawa tawa last updated on 02/Jun/17

Determine the fourth roots of - 16, giving the results in the form a + jb .

Answered by mrW1 last updated on 03/Jun/17

z^{4} = - 16

z = x + i y = r (\cos θ + i \sin θ)

z^{4} = r^{4} (\cos 4 θ + i \sin 4 θ) = - 16 = 16 (- 1 + i 0)

\Rightarrow r^{4} = 16 \Rightarrow r = 2

\Rightarrow \cos 4 θ = - 1

\Rightarrow \sin 4 θ = 0

\Rightarrow 4 θ = (2 n - 1) π, n = 1, 2, 3, 4

θ_{1} = \frac{π}{4}

θ_{2} = \frac{3 π}{4}

θ_{3} = \frac{5 π}{4}

θ_{4} = \frac{7 π}{4}

z_{1} = 2 (\cos \frac{π}{4} + i \sin \frac{π}{4}) = \sqrt{2} + i \sqrt{2}

z_{2} = 2 (\cos \frac{3 π}{4} + i \sin \frac{3 π}{4}) = - \sqrt{2} + i \sqrt{2}

z_{3} = 2 (\cos \frac{5 π}{4} + i \sin \frac{5 π}{4}) = - \sqrt{2} - i \sqrt{2}

z_{4} = 2 (\cos \frac{7 π}{4} + i \sin \frac{7 π}{4}) = \sqrt{2} - i \sqrt{2}

Commented by tawa tawa last updated on 04/Jun/17

Wow, i really appreciate sir . God bless you .

Commented by RasheedSoomro last updated on 04/Jun/17

\in \times cellent!