Determine-the-fourth-roots-of-16-giving-the-results-in-polar-form-and-in-exponential-form-Answers-2-1-j-2-1-j-2-1-j-2-1-j-

Question Number 50806 by Tawa1 last updated on 20/Dec/18

Determine the fourth roots of - 16, giving the results in polar

form and in exponential form

Answers : \sqrt{2} (1 + j), \sqrt{2} (- 1 + j), \sqrt{2} (- 1 - j), \sqrt{2} (1 - j)

Answered by mr W last updated on 20/Dec/18

x = r (\cos θ + j \sin θ)

x^{4} = - 16

r^{4} (\cos 4 θ + j \sin 4 θ) = 2^{4} (- 1 + 0 j)

\Rightarrow r = 2

\cos 4 θ = - 1

\sin 4 θ = 0

\Rightarrow 4 θ = π, 3 π, 5 π, 7 π

\Rightarrow θ = \frac{π}{4}, \frac{3 π}{4}, \frac{5 π}{4}, \frac{7 π}{4}

x = 2 (\cos \frac{π}{4} + j \sin \frac{π}{4}) = \sqrt{2} (1 + j)

x = 2 (\cos \frac{3 π}{4} + j \sin \frac{3 π}{4}) = \sqrt{2} (- 1 + j)

x = 2 (\cos \frac{5 π}{4} + j \sin \frac{5 π}{4}) = \sqrt{2} (1 - j)

x = 2 (\cos \frac{7 π}{4} + j \sin \frac{7 π}{4}) = \sqrt{2} (- 1 - j)

Commented by Tawa1 last updated on 20/Dec/18

God bless you sir

Commented by Tawa1 last updated on 21/Dec/18

Sir, am trying to understand how you got 3 π, 5 π, 7 π \dots

I thought the angles should be in the same interval sir . like in

interval of \frac{360}{4} = 90 since it is 4 th roots .

so, 4 θ = 180, 270, 360, 450, etc \dots

Please sir, i want to know why we use π, 3 π, 5 π, 7 π, and not the

interval of 4 th roots \frac{360}{4} = 90 .

God bless you sir

Commented by mr W last updated on 21/Dec/18

0 ⩽ θ ⩽ 2 π

\Rightarrow 0 ⩽ 4 θ ⩽ 8 π

i n t h i s r a n g e w e h a v e f o l l o w i n g v a l u e s

f o r 4 θ : π, 3 π, 5 π, 7 π

w h i c h f u l f i l l

\cos 4 θ = - 1

\sin 4 θ = 0

Answered by peter frank last updated on 21/Dec/18

z^{4} = - 16 (\cos π + j \sin π)

r = \sqrt{16^{2}} = 16

f r o m

z_{k} = r^{\frac{1}{n}} [\cos \frac{θ + 2 π k}{n} + j \sin \frac{θ + 2 π k}{n}]

n = 4 θ = π

z_{k} = 2 [\cos \frac{π + 2 π k}{4} + j \sin \frac{π + 2 π k}{4}]

k = 0

z_{0} = 2 (\cos \frac{π}{4} + j \sin \frac{π}{4}) = \sqrt{2} (1 + j)

k = 1

z_{1} = 2 (\cos \frac{3 π}{4} + j \sin \frac{3 π}{4}) = - \sqrt{2} (1 - j)

k = 2

z_{2} = 2 (\cos \frac{5 π}{4} + j \sin \frac{5 π}{4}) = \sqrt{2} (- 1 - j)

k = 3

z_{3} = 2 (\cos \frac{7 π}{4} + j \sin \frac{7 π}{4}) = \sqrt{2} (1 - j)

Commented by Tawa1 last updated on 21/Dec/18

God bless you sir

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