Find complex roots of $\frac{2i}{1+i}$ 
Find 6th roots of $$\frac{2i}{1+i}$$

$$\frac{2i}{1+i}=\frac{2e^{i\pi/2}}{\sqrt 2 e^{i \pi/4}}=\sqrt 2 e^{i \pi/4}$$
Now if I set $z^{1/6}=\sqrt 2 e^{i \pi/4}$ and knowing the fact that the roots are distibuted equality with an angle $k\pi/3$ for $k=1,2,3,4,5,6$ I get the answer to be:
$$2^{1/12}( \cos(\pi/24 + k\pi/3)+i\sin(\pi/24 + k\pi/3))$$
Is this correct?
 A: you are solving $$z^6=\sqrt{2}e^{i(\frac{\pi}{4}+k.2\pi)}$$therefore$$z=2^{\frac{1}{12}}\left(\cos(\frac{\pi}{24}+k\frac{\pi}{3})+i\sin(\frac{\pi}{24}+k\frac{\pi}{3})\right)$$
A: $$z^6=\frac{2i}{1+i}\Longleftrightarrow$$
$$z^6=\left|\frac{2i}{1+i}\right|e^{\arg\left(\frac{2i}{1+i}\right)i}\Longleftrightarrow$$
$$z^6=\frac{\left|2i\right|}{\left|1+i\right|}e^{\left(\arg(2i)-\arg(1+i)\right)i}\Longleftrightarrow$$
$$z^6=\frac{2}{\sqrt{2}}e^{\left(\frac{\pi}{2}-\arctan\left(\frac{1}{1}\right)\right)i}\Longleftrightarrow$$
$$z^6=\sqrt{2}e^{\left(\frac{\pi}{2}-\frac{\pi}{4}\right)i}\Longleftrightarrow$$
$$z^6=\sqrt{2}e^{\frac{\pi}{4}i}\Longleftrightarrow$$
$$z=\left(\sqrt{2}e^{\left(2\pi k+\frac{\pi}{4}\right)i}\right)^{\frac{1}{6}}\Longleftrightarrow$$
$$z=\sqrt[12]{2}e^{\frac{1}{6}\left(2\pi k+\frac{\pi}{4}\right)i}$$
With $k\in\mathbb{Z}$ and $k:0-5$.

So the solutions are:
$$z_0=\sqrt[12]{2}e^{\frac{1}{6}\left(2\pi\cdot0+\frac{\pi}{4}\right)i}=\sqrt[12]{2}e^{\frac{\pi}{24}i}$$
$$z_1=\sqrt[12]{2}e^{\frac{1}{6}\left(2\pi\cdot1+\frac{\pi}{4}\right)i}=\sqrt[12]{2}e^{\frac{3\pi}{8}i}$$
$$z_2=\sqrt[12]{2}e^{\frac{1}{6}\left(2\pi\cdot2+\frac{\pi}{4}\right)i}=\sqrt[12]{2}e^{\frac{17\pi}{24}i}$$
$$z_3=\sqrt[12]{2}e^{\frac{1}{6}\left(2\pi\cdot3+\frac{\pi}{4}\right)i}=\sqrt[12]{2}e^{-\frac{23\pi}{24}i}$$
$$z_4=\sqrt[12]{2}e^{\frac{1}{6}\left(2\pi\cdot4+\frac{\pi}{4}\right)i}=\sqrt[12]{2}e^{-\frac{5\pi}{8}i}$$
$$z_5=\sqrt[12]{2}e^{\frac{1}{6}\left(2\pi\cdot5+\frac{\pi}{4}\right)i}=\sqrt[12]{2}e^{-\frac{7\pi}{24}i}$$
