Calculating the residues at the poles of $f(z) = \frac{\tan(z) }{z^2 + z +1} $ 

No solution as discussed in comment- Aman R

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" Let us first find the poles: 
\[z^2+z+1=0  \Rightarrow z=\frac{-1\pm\sqrt{3}i}{2}=-\frac{1}{2}\pm \frac{\sqrt{3}}{2}i.\]
Let $z_1=-\frac{1}{2}+ \frac{\sqrt{3}}{2}i$ and $z_2=-\frac{1}{2}- \frac{\sqrt{3}}{2}i.$ Then $z^2+z+1= (z-z_1)(z-z_2)$, and

\[ \text{Res}(f, z_1)=\lim_{z \rightarrow z_1} (z- z_1)\frac{\tan (z)}{z^2+z+1} =\frac{\tan (z_1)}{z_1-z_2} \]
\[=\frac{\tan (-\frac{1}{2}+ \frac{\sqrt{3}}{2}i)}{\sqrt{3}i}\]
\[=\frac{\frac{\sin(-1)+i\sinh (\sqrt{3})}{\cos(-1)+i\cosh(\sqrt{3})}}{\sqrt{3}i}=\frac{\sin(-1)+i\sinh (\sqrt{3})}{\sqrt{3}\cos(-1)i-\sqrt{3}\cosh(\sqrt{3})}.\]

Similarly 
\[ \text{Res}(f, z_2)=\lim_{z \rightarrow z_2} (z- z_2)\frac{\tan (z)}{z^2+z+1} =\frac{\tan (z_2)}{z_2-z_1} \] \[=\frac{\tan (-\frac{1}{2}- \frac{\sqrt{3}}{2}i)}{-\sqrt{3}i}\] \[=\frac{\frac{\sin(-1)+i\sinh (-\sqrt{3})}{\cos(-1)+i\cosh(-\sqrt{3})}}{-\sqrt{3}i}=\frac{\sin(-1)+i\sinh (-\sqrt{3})}{-\sqrt{3}\cos(-1)i+\sqrt{3}\cosh(\sqrt{3})}.\]

You may rationalize the final answers to get a compex number of the form $a+bi$, but I beilive the above form is much simpler. 

Here we are using the formula
\[\tan(a+bi)=\frac{\sin(2a)+i\sinh(2b)}{\cos(2a)+i\cosh(2b)}.\]"- Philip