what are the fourth roots of z1= √ 3+i ?

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You need to evaluate the roots of `z = sqrt(3 + i)` , hence, you should use De Moivre's theorem, but first you need to convert the algebraic form of complex number `z' = 3 + i ` ,under the square root, into polar form, such that:

`|z'| = sqrt(3^2 + 1^2) => |z'| = sqrt 4 => |z'| = 2`

`tan alpha = 1/3 => alpha = tan^(-1)(1/3) ~~ 18^o = pi/10`

`z' = 2(cos (pi/10) + i sin(pi/10)) = > z = sqrt(z') => z = (z')^(1/2)`

`z = (2(cos (pi/10) + i sin(pi/10)))^(1/2)`

Using De Moivre's theorem yields:

`z = sqrt 2(cos (pi/10 + 2kpi)/2 + i sin (pi/10 + 2kpi)/2)`

Substituting `k = 0,1,2,3` yields:

`z_1 = sqrt 2(cos (pi/20) + i sin (pi/20))`

`z_2 = sqrt2(cos (pi/10 + 2pi)/2 + i sin (pi/10 + 2pi)/2)`

`z_2 = sqrt2(cos (21pi/20) + i sin (21pi/20))`

`z_3 = sqrt2(cos (pi/10 + 4pi)/2 + i sin (pi/10 + 4pi)/2)`

`z_3 = sqrt2(cos(41pi/20) + i sin (41pi/20))`

`z_4 = sqrt2(cos (pi/10 + 6pi)/2 + i sin (pi/10 + 6pi)/2)`

`z_4 = sqrt2(cos (61pi/20) + i sin (61pi/20))`

Hence, evaluating all the four roots of the given equation yields

` z_1 = sqrt 2(cos (pi/20) + i sin (pi/20)) ; z_2 = sqrt2(cos (21pi/20) + i sin (21pi/20)) ; z_3 = sqrt2(cos (41pi/20) + i sin (41pi/20)) ; z_4 = sqrt2(cos (61pi/20) + i sin (61pi/20)). `

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