Simplify by hand!

[Deleted User]

square root (2/square root (3) - 2i) where i is imaginary number

[b^3]

Hint: Try 'rationalising' the denominator first.

Spoiler

 

[Phy124]

Sorry the -2i part isn't part of the division.
What I meant was sqrt [ (2/sqrt(3)) - 2i ]   

We know that:



Therefore using de Moivre's formula we can show that;

[silverpixeli]

Am I wrong in saying that (some imaginary number)^1/2 should have two solutions? Or do we only take the principal solution in this case because the original question asked for the sqrt() which is the principal square root?

[Phy124]

Am I wrong in saying that (some imaginary number)^1/2 should have two solutions? Or do we only take the principal solution in this case because the original question asked for the sqrt() which is the principal square root?

I only took the principal root, but I wasn't entirely sure whether I should have given the second as well. Having not done spesh, I'm not sure whether it assumed you list all roots or list the principal and list the others when asked to do so.

I think that it is the latter but on second thought, it could be the former, where you list all solutions of the domain [-pi,pi], in which case the second root will be the negative of the original, i.e.



tl;dr: I can't answer with confidence because I didn't do the subject, best to ask your teacher

edit: thanks for clearing that up abcdqd 

[abcdqd]

if the question asks for the sqrt, you have to reject the negative answer
if it just says find the roots of, you have to list all of them

[Deleted User]

Thank you!