Answer
$-\displaystyle \frac{\pi}{8}.\quad $
(converges)
Work Step by Step
$\displaystyle \int\frac{z}{z^{4}+4}dz=\left[\begin{array}{l}
u=z^{2}\\
du=2zdz
\end{array}\right]=\frac{1}{2}\int\frac{du}{u^{2}+2^{2}}du$
Table of integrals: 17.
$\displaystyle \int\frac{du}{a^{2}+u^{2}}=\frac{1}{a}\tan^{-1}\frac{u}{a}+C$
The integral is type I improper integral, so
$\displaystyle \int_{-\infty}^{0}\frac{z}{z^{4}+4}dz=\lim_{t\rightarrow-\infty}\int_{t}^{0}\frac{z}{z^{4}+4}dz=\lim_{t\rightarrow-\infty}\frac{1}{2}\left[\frac{1}{2}\tan^{-1}\left(\frac{z^{2}}{2}\right)\right]_{t}^{0}\quad $
$=\displaystyle \lim_{t\rightarrow-\infty}\left[0-\frac{1}{4}\tan^{-1}\left(\frac{t^{2}}{2}\right)\right]$
$=-\displaystyle \frac{1}{4}\left(\frac{\pi}{2}\right)$
$=-\displaystyle \frac{\pi}{8}.\quad $
(converges)
