Principal nth root

[Aladdin]

(*) Give the principal nth root of :


\(\displaystyle \frac{(2-2i\sqrt3)^4(-1-i)^2}{(\sqrt3-i)^3}\)\(\displaystyle . \ . All \ to \ the \ power \ of \ 1/3.\)

I've used many methods but all where too long.


Thanks for any help

 

[galactus]

Inside the 1/3 power, it whittles down to \(\displaystyle 32-32\sqrt{3}i\)

Then, \(\displaystyle \sqrt[3]{32-32\sqrt{3}i}=4cos(\frac{\pi}{9})-4i\cdot sin(\frac{\pi}{9})\)

Use DeMoivre's Theorem

\(\displaystyle \sqrt[n]{r}\left[cos\left(\frac{{\theta}+2\pi k}{n}\right)+i\cdot sin\left(\frac{{\theta}+2{\pi}k}{n}\right)\right]\)

Let k=0 and n=3

Note that \(\displaystyle r=\sqrt{(32)^{2}+(32\sqrt{3})^{2}}=64\)

So, \(\displaystyle \sqrt[3]{64}=4\)

 

[Deleted member 4993]

(*) Give the principal nth root of :


\(\displaystyle \frac{(2-2i\sqrt3)^4(-1-i)^2}{(\sqrt3-i)^3}\)\(\displaystyle . \ . All \ to \ the \ power \ of \ 1/3.\)

I've used many methods but all where too long.


Thanks for any help

\(\displaystyle \left [\frac{4^4(\frac{1}{2}-i\frac{\sqrt{3}}{2})^4\cdot (\sqrt{2})^2( -\frac{1}{\sqrt{2}}-i\frac{1}{\sqrt{2}})^2}{(2)^3(\frac{\sqrt{3}}{2}-i\frac{1}{2})^3}\right]^{\frac{1}{3}}\)

\(\displaystyle \text principal \ \ root \ = \ \left [\frac{4^4\cdot (\sqrt{2})^2}{(2)^3}\right]^{\frac{1}{3}} \ = \ 4\)

 

[Aladdin]

printf("Thanks for the help , guys.\n");