[tex]\bf \sqrt[{{ n}}]{z}=\sqrt[{{ n}}]{r}\left[ cos\left( \frac{\theta+2\pi k}{{{ n}}} \right) +i\ sin\left( \frac{\theta+2\pi k}{{{ n}}} \right)\right]\quad k\ roots\\\\
-------------------------------\\\\
\sqrt[{{ 4}}]{z}=\sqrt[{{ 4}}]{r}\left[ cos\left( \frac{\theta+2\pi k}{{{ 4}}} \right) +i\ sin\left( \frac{\theta+2\pi k}{{{ 4}}} \right)\right]\quad
\begin{array}{llll}
k\ roots\\
0,1,2,3
\end{array}\\\\
-------------------------------\\\\[/tex]
[tex]\bf \sqrt[{{ 4}}]{256}\left[ cos\left( \frac{280^o+360^o(0)}{{{ 4}}} \right) +i\ sin\left( \frac{280^o+360^o(0)}{{{ 4}}} \right)\right]
\\\\\\
4[cos(70^o)+i\ sin(70^o)]\implies 4cos(70^o)+i\ 4sin(70^o)\impliedby
\begin{array}{llll}
1st\ root\\
k=0
\end{array}[/tex]
[tex]\bf -------------------------------\\\\
\sqrt[{{ 4}}]{256}\left[ cos\left( \frac{280^o+360^o(1)}{{{ 4}}} \right) +i\ sin\left( \frac{280^o+360^o(1)}{{{ 4}}} \right)\right]
\\\\\\
4[cos(160^o)+i\ sin(160^o)]
\\\\\\
4cos(160^o)+i\ 4sin(160^o)\impliedby
\begin{array}{llll}
2nd\ root\\
k=1
\end{array}[/tex]
[tex]\bf -------------------------------\\\\
\sqrt[{{ 4}}]{256}\left[ cos\left( \frac{280^o+360^o(2)}{{{ 4}}} \right) +i\ sin\left( \frac{280^o+360^o(2)}{{{ 4}}} \right)\right]
\\\\\\
4[cos(250^o)+i\ sin(250^o)]
\\\\\\
4cos(250^o)+i\ 4sin(250^o)\impliedby
\begin{array}{llll}
3rd\ root\\
k=2
\end{array}[/tex]
[tex]\bf -------------------------------\\\\
\sqrt[{{ 4}}]{256}\left[ cos\left( \frac{280^o+360^o(3)}{{{ 4}}} \right) +i\ sin\left( \frac{280^o+360^o(3)}{{{ 4}}} \right)\right]
\\\\\\
4[cos(340^o)+i\ sin(340^o)]
\\\\\\
4cos(340^o)+i\ 4sin(340^o)\impliedby
\begin{array}{llll}
4th\ root\\
k=3
\end{array}[/tex]
so.. just get their product, for example the first one would be 1.37 + 3.76i rounded up
