How can I calculate the square root of&nbsp;5(cos⁡120+isin⁡120)?

Use&nbsp;cos⁡(180∘+a)=−cos⁡a&nbsp;and&nbsp;sin⁡(180∘+a)=−sin⁡a&nbsp;(5(cos⁡120∘+isin⁡120∘))12&nbsp;5(cos⁡(2k(180)∘+120∘)+isin⁡(2k(180)∘+120∘))12&nbsp;=5(cos⁡(2k(180)+120∘2)+isin⁡((2k(180)∘+120∘)2))×k=0,&nbsp;1&nbsp;=5(cos⁡60∘+isin⁡60∘)=52(1+i3), for&nbsp;k=0&nbsp;and&nbsp;=5(cos⁡240∘+isin⁡240∘)&nbsp;for&nbsp;k=1&nbsp;=5(cos⁡(180∘+60∘)+isin⁡(180∘+60∘)&nbsp;&nbsp;=−5(cos⁡60∘+i∈60∘)&nbsp;=−52(1+i3)

"How can I calculate the square root of5(cos⁡120+isin⁡120)?" was answered by our experts

trefoniu1

Answered question

2022-02-01

How can I calculate the square root of $5 (\cos 120 + i \sin 120) ?$

Answer & Explanation

becky4208fj

Beginner2022-02-02Added 10 answers

Use $\cos (180^{\circ} + a) = - \cos a$ and $\sin (180^{\circ} + a) = - \sin a$
${(5 ({\cos 120}^{\circ} + i {\sin 120}^{\circ}))}^{\frac{1}{2}}$
$\sqrt{5} {(\cos (2 k {(180)}^{\circ} + 120^{\circ}) + i \sin (2 k {(180)}^{\circ} + 120^{\circ}))}^{\frac{1}{2}}$
$= \sqrt{5} (\cos (\frac{2 k (180) + 120^{\circ}}{2}) + i \sin (\frac{(2 k {(180)}^{\circ} + 120^{\circ})}{2})) \times k = 0, 1$
$= \sqrt{5} ({\cos 60}^{\circ} + i {\sin 60}^{\circ}) = \frac{\sqrt{5}}{2} (1 + i \sqrt{3})$ , for $k = 0$ and
$= \sqrt{5} ({\cos 240}^{\circ} + i {\sin 240}^{\circ})$ for $k = 1$
$= \sqrt{5} (\cos (180^{\circ} + 60^{\circ}) + i \sin (180^{\circ} + 60^{\circ})$
$= - \sqrt{5} ({\cos 60}^{\circ} + i \in 60^{\circ})$
$= - \frac{\sqrt{5}}{2} (1 + i \sqrt{3})$

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