When do you take into account the +2kpi for complex numbers arguments in complex equations $$z^2 = ({2e^{i{\frac{\pi}{3}}}})^8$$
To find z I took the square root of both sides which gives me:
$$z = ({2e^{i{\frac{\pi}{3}}}})^4$$ which I rewrote as $$z = {2^4e^{i{\frac{4\pi}{3}}+2k\pi}}$$
and for k=0 we find $$z=-8-8\sqrt(3)$$ which is one of the correct answers but for k=1 we find the same thing.
On the other hand the correction states the other answer is $$z=8+8\sqrt(3)$$
I think that I need to put the $$+2k\pi$$ before taking the square root of Z but I don't understand why it changes the answer since mathematically it should be the same?
Because even when I tried with 
$$z = ({2e^{i{\frac{\pi}{3}}+2k\pi}})^4$$ which I rewrote as $$z = {2^4e^{i{\frac{4\pi}{3}}+8k\pi}}$$
I still don't find the correct second answer.
Thanks

*

*You can use this method:


$$z^2=(2e^{i{\pi\over3}})^8=(2^8e^{i{8\pi\over3}})=(2^8e^{i{8\pi\over3}+2ki\pi})$$
and now you take the square root.


*Otherwise you can take the square root putting $\pm$:


$$z=\pm(2e^{i{\pi\over3}})^4$$


*We can also write $2e^{i{\pi\over3}}=2e^{i{\pi\over3}+2li\pi}$ because any complex number is defined up to a phase factor $e^{2li\pi}$, so:


$$z^2=(2e^{i{\pi\over3}+2li\pi})^8=(2^8e^{i{8\pi\over3}+16li\pi})=(2^8e^{i{8\pi\over3}+16li\pi+2ki\pi})=(2^8e^{i{8\pi\over3}+2\pi i(8l+k)})$$
$$z=2^4e^{i{4\pi\over3}+\pi i(8l+k)}=2^4e^{i{4\pi\over3}}e^{\pi i(8l+k)}=\pm2^4e^{i{4\pi\over3}}$$
with $l,k\in\mathbb Z$, but $8l+k$ can be even or odd so you get the same solutions.
