Residue calculations involving logarithms. I am trying to solve a definite integral using residue theory. At some step, I wish to calculate the residue of the function $\frac{\log{(z)}}{z^3+8}$ at $-2e^{2\pi i/3}$. The log function is defined for $0 < \arg z < 2\pi$.
I'm having difficulty doing this simple looking calculation. Can someone please help me with this? 
I have been given a hint that $Res(\frac{\log{(z)}}{z^3+8},z_i)=\frac{-z_i log(z_i)}{24}$. Why and how is this true?
 A: If you are having a problem computing the value of $\log z$ at that value:
Note that we have, $$\log z = \log |z| +i\theta, \, \, -\pi < \theta \leq \pi$$
Now, we have, $z=-2e^{\frac{2\pi i}{3}}=-2e^{-\frac{\pi i}{3}}$, giving us, $|z|=2$ and $\theta= -\frac{\pi}{3}$. Now, use the formula.

Edit:
Now, let us calculate the residue at $z=-2e^{\frac{2\pi i}{3}}= 1-\sqrt{3}i$. We have, $$\text{Res } f(z)_{z = -2e^{\frac{2\pi i}{3}}} = \frac{\log (-2e^{\frac{2\pi i}{3}})}{((1-\sqrt{3}i)+2)((1-\sqrt{3}i)-(1+\sqrt{3}i))}$$ $$=\frac{\log 2 - \frac{\pi i}{3}}{(3-\sqrt{3}i)(-2\sqrt{3}i)}$$ $$=[\log 2 - i\frac{\pi}{3}]\frac{1}{-6-6\sqrt{3}i}$$ $$=[\log 2 -i\frac{\pi}{3}]\frac{-6+6\sqrt{3}i}{144}$$ $$=-\frac{1-\sqrt{3}i}{24}[\log 2 -i\frac{\pi}{3}]$$ $$=-\frac{z\log z}{24}\bigg\lvert_{z=-2e^{\frac{2\pi i}{3}}}$$
A: $$\frac{\log(z)}{z^3+8}= \frac{\log(z)}{(z+2e^{\frac{2i\pi}3})(z^2-2e^{\frac{2i\pi}{3}}z+4e^{\frac{-2i\pi}{3}})}$$
Then the residue at $\displaystyle z=-2e^{\frac{2\pi i}{3}}$ should be:
$$\frac{\log(-2e^{\frac{2\pi i}{3}})}{(-2e^{\frac{2\pi i}{3}})^2-2(-2e^{\frac{2\pi i}{3}})(e^{\frac{2i\pi}{3}})+4e^{\frac{-2i\pi}{3}}}= \frac{\log (-2e^{\frac{2i\pi}{3}})}{4e^{\frac{4i\pi}{3}}+ 4e^{\frac{4i\pi}{3}}+ 4e^{\frac{-2i\pi}{3}}}= \frac{\log( -2e^{\frac{2i\pi}{3}})}{8e^{\frac{4i\pi}{3}}+ 4e^{\frac{-2i\pi}{3}}}=  \frac{e^{\frac{2i\pi}{3}}\log( -2e^{\frac{2i\pi}{3}})}{8e^{\frac{6i\pi}{3}}+ 4}=\frac{e^{\frac{2i\pi}{3}}\log( -2e^{\frac{2i\pi}{3}})}{12}$$
