Question

Which expression is a fourth root of -1+isqrt3?

Answer 1

Explanation:

`\sf n^(th)` roots of a complex number is given by DeMoivre's formula.

`\sf \boxed{\bf r^{(1)/(n)}\left[Cos (\theta + 2\pi k)/(n)+i \ Sin \ (\theta+2\pi k)/(n)\right]}`

Here, k lies between 0 and (n -1) ; n is the exponent.

`\sf -1 + i√(3)`

a = -1 and b = √3

`\sf \boxed{r=√(a^2+b^2)} \ and \ \boxed{\theta = Tan^(-1) \ (b)/(a)}`

`\sf r = √((-1)^2 + 3^2)\\\\ = √(1+9)\\\\=√(10)`

`\sf \theta = tan^(-1) \ (√(3))/(-1)\\\\ = tan^(-1) \ (-√(3))`

`\sf = (-\pi )/(3)`

n = 4

For k = 0,

`\sf z = \sqrt[4]{10}\left[Cos \ ((-\pi)/(3) +0)/(4)+iSin \ ((-\pi)/(3)+0)/(4)\right] \\\\\\z= \sqrt[4]{10} \left[Cos \ ( -\pi )/(12)+iSin \ (-\pi)/(12)\right]\\\\\\z = \sqrt[4]{10}\left[-Cos \ (\pi)/(12)-i \ Sin \ (\pi)/(12)\right]`

For k =1,

`\sf z = \sqrt[4]{10}\left[Cos \ (5\pi)/(12)+i \ Sin \ (5\pi)/(12)\right]`

For k =2,

`z = \sqrt[4]{10}\left[Cos \ (11\pi)/(12)+i \ Sin \ (11\pi)/(12)\right]`

For k = 3,

`\sf z = \sqrt[4]{10}\left[Cos \ (17\pi)/(12)+i \ Sin \ (17\pi)/(12)\right]`

For k = 4,

`\sf z =\sqrt[4]{10}\left[Cos \ (23\pi)/(12)+i \ Sin \ (23\pi)/(12)\right]`