# How do you find the 6th root of #sqrt3-i #?

##### 1 Answer

Let

First let us plot the point

And we will put the complex number into polar form:

# |omega| = sqrt(3+1) = 2 #

# arg(omega) = tan^-1(-1/sqrt(3)) = -(pi)/6 #

So then in polar form we have:

# omega = 2(cos(-(pi)/6) + isin(-(pi)/6)) #

We now want to solve the equation for

# z^6 = 2(cos(-(pi)/6) + isin(-(pi)/6)) #

Whenever dealing with complex variable equation such as this it is essential to remember that the complex exponential has a period of

# z^6 = 2(cos(2npi-(pi)/6) + isin(2npi-(pi)/6)) \ \ \ n in NN#

By De Moivre's Theorem we can write this as:

# z = {2(cos(2npi-(pi)/6) + isin(2npi-(pi)/6))}^(1/6)#

# \ \ = 2^(1/6)(cos((2npi-(pi)/6)/6) + isin((2npi-(pi)/6)/6))#

# \ \ = 2^(1/6)(cos theta + isin theta) \ \ \ \ # where# theta=((12n-1)pi)/36#

Put:

# n=1 => theta = (11pi)/36 => z = 2^(1/6)(cos ((11pi)/36)+ isin ((11pi)/36)) #

# n=2 => theta = (23pi)/36 => z = 2^(1/6)(cos ((23pi)/36)+ isin ((23pi)/36)) #

# n=3 => theta = (35pi)/36=> z = 2^(1/6)(cos ((35pi)/36)+ isin ((35pi)/36)) #

# n=4 => theta = (47pi)/36=> z = 2^(1/6)(cos ((47pi)/36)+ isin ((47pi)/36)) #

# n=5 => theta = (59pi)/36=> z = 2^(1/6)(cos ((59pi)/36)+ isin ((59pi)/36)) #

# n=6 => theta = (71pi)/36=> z = 2^(1/6)(cos ((71pi)/36)+ isin ((71pi)/36)) #

After which the pattern continues.

We can plot these solutions on the Argand Diagram