How do you solve $2 x^{4} + 2 + 2 \sqrt{3} i = 0$ $2x^4+2+2sqrt3i=0$ ?

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How do you solve $2 x^{4} + 2 + 2 \sqrt{3} i = 0$ $2x^4+2+2sqrt3i=0$ ?

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Answer 1 · 2017-02-04T07:31:03

$x = 2^{\frac{1}{4}} (cos (\frac{5 π}{12}) + i sin (\frac{5 π}{12}))$ $x=2^(1/4)(cos((5pi)/12)+isin((5pi)/12))$ or $2^{\frac{1}{4}} (- cos (\frac{π}{12}) + i sin (\frac{π}{12}))$ $2^(1/4)(-cos((pi)/12)+isin((pi)/12))$ or $2^{\frac{1}{4}} (- cos (\frac{5 π}{12}) - i sin (\frac{5 π}{12}))$ $2^(1/4)(-cos((5pi)/12)-isin((5pi)/12))$ or $2^{\frac{1}{4}} (cos (\frac{π}{12}) - i sin (\frac{π}{12}))$ $2^(1/4)(cos((pi)/12)-isin((pi)/12))$

Explanation:

As $2 x^{4} + 2 + 2 \sqrt{3} i = 0$ $2x^4+2+2sqrt3i=0$

$x^{4} = - 1 - \sqrt{3} i$ $x^4=-1-sqrt3i$

Hence $x = {(- 1 - \sqrt{3} i)}^{\frac{1}{4}}$ $x=(-1-sqrt3i)^(1/4)$

To find this we intend to use DeMoivres Theorem. So let us convert $- 1 - \sqrt{3} i$ $-1-sqrt3i$ in polar form.

As $∣ ∣ - 1 - \sqrt{3} i ∣ ∣ = 2$ $|-1-sqrt3i|=2$ , $- 1 - \sqrt{3} i = 2 (- \frac{1}{2} - \frac{\sqrt{3}}{2} i)$ $-1-sqrt3i=2(-1/2-sqrt3/2i)$

or $- 1 - \sqrt{3} i = 2 (cos (2 n π + \frac{5 π}{3}) + i sin (2 n π + \frac{5 π}{3}))$ $-1-sqrt3i=2(cos(2npi+(5pi)/3)+isin(2npi+(5pi)/3))$

and $x = {(- 1 - \sqrt{3} i)}^{\frac{1}{4}} = 2^{\frac{1}{4}} (cos (\frac{2 n π}{4} + \frac{5 π}{12}) + i sin (\frac{2 n π}{4} + \frac{5 π}{12}))$ $x=(-1-sqrt3i)^(1/4)=2^(1/4)(cos((2npi)/4+(5pi)/12)+isin((2npi)/4+(5pi)/12))$

= $2^{\frac{1}{4}} (cos (\frac{n π}{2} + \frac{5 π}{12}) + i sin (\frac{n π}{2} + \frac{5 π}{12}))$ $2^(1/4)(cos((npi)/2+(5pi)/12)+isin((npi)/2+(5pi)/12))$

Now choosing four values of $n$ $n$ i.e. $0, 1, 2$ $0,1,2$ and $3$ $3$ , we can get four roots of the equation $2 x^{4} + 2 + 2 \sqrt{3} i = 0$ $2x^4+2+2sqrt3i=0$ , which are

$2^{\frac{1}{4}} (cos (\frac{5 π}{12}) + i sin (\frac{5 π}{12}))$ $2^(1/4)(cos((5pi)/12)+isin((5pi)/12))$ ,

$2^{\frac{1}{4}} (cos (\frac{11 π}{12}) + i sin (\frac{11 π}{12}))$ $2^(1/4)(cos((11pi)/12)+isin((11pi)/12))$ i.e. $2^{\frac{1}{4}} (- cos (\frac{π}{12}) + i sin (\frac{π}{12}))$ $2^(1/4)(-cos((pi)/12)+isin((pi)/12))$

$2^{\frac{1}{4}} (cos (\frac{17 π}{12}) + i sin (\frac{17 π}{12}))$ $2^(1/4)(cos((17pi)/12)+isin((17pi)/12))$ i.e. $2^{\frac{1}{4}} (- cos (\frac{5 π}{12}) - i sin (\frac{5 π}{12}))$ $2^(1/4)(-cos((5pi)/12)-isin((5pi)/12))$

and $2^{\frac{1}{4}} (cos (\frac{23 π}{12}) + i sin (\frac{23 π}{12}))$ $2^(1/4)(cos((23pi)/12)+isin((23pi)/12))$ i.e. $2^{\frac{1}{4}} (cos (\frac{π}{12}) - i sin (\frac{π}{12}))$ $2^(1/4)(cos((pi)/12)-isin((pi)/12))$

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How do you solve $2 x^{4} + 2 + 2 \sqrt{3} i = 0$ $2x^4+2+2sqrt3i=0$ ?

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Explanation:

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