234k views
2 votes
Which expression is a fourth root of -1+isqrt3?

1 Answer

1 vote

Answer:

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]

User Oersted
by
9.1k points

Related questions

asked Oct 26, 2024 209k views
Runofthemillgeek asked Oct 26, 2024
by Runofthemillgeek
7.6k points
1 answer
4 votes
209k views
asked Dec 20, 2020 190k views
Mark Omo asked Dec 20, 2020
by Mark Omo
7.5k points
1 answer
0 votes
190k views
asked Oct 23, 2021 48.2k views
Enricog asked Oct 23, 2021
by Enricog
8.0k points
1 answer
4 votes
48.2k views
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.