How Do You Calculate Cube Roots of Complex Numbers in Modulus Argument Form?

[whooi]

can someone help me solve this question as fasd as possible??
im really headache wif it...
thx~~

determine the 3 cube roots of 3-i over 3+i giving the result in modulus argument form,express the principal root in the form a+Jb

 

[whooi]

hello..anyone can help me out??

 

[berkeman]

can someone help me solve this question as fasd as possible??
im really headache wif it...
thx~~

determine the 3 cube roots of 3-i over 3+i giving the result in modulus argument form,express the principal root in the form a+Jb

Do not bump your post after only 16 minutes. It is unreasonable to expect fast help all the time here on the PF.

Also, you must show your attempt at a solution before we can be of help. Show us how you would go about simplifying the complex fraction that you are describing...

 

[Mark44]

First off, you're going to want to calculate (3 - i)/(3 + i), to get it in the form a + bi. After that, you should calculate the polar form of this complex number, r(cos$\theta$ + i sin$\theta$). After that, you can use DeMoivre's Formula to find a cube root, which says that
$$(r(cos\theta + i sin\theta))^{1/n} = r^{1/n}(cos(\theta/n) + i sin(\theta/n))$$.