$(\sqrt{3}-\mathrm{i}{)}^{2016}+(-\sqrt{3}-\mathrm{i}{)}^{2019}$

$=2^{2019}\left\{{\left(\frac{\sqrt{3}}{2}-\frac{\mathrm{i}}{2}\right)}^{2016}\frac{1}{8}+{\left(-\frac{\sqrt{3}}{2}-\frac{\mathrm{i}}{2}\right)}^{2019}\right\}$

$=2^{2019}\left\{{\left(\cos \left(-\frac{\pi }{6}\right)+i\sin \left(-\frac{\pi }{6}\right)\right)}^{2016}\frac{1}{8}+{\left(\cos \left(\frac{7\pi }{6}\right)+i\sin \left(\frac{7\pi }{6}\right)\right)}^{2019}\right\}$

$=2^{2019}\left\{{\left(\cos \left(\frac{11\pi }{6}\right)+i\sin \left(\frac{11\pi }{6}\right)\right)}^{2016}\frac{1}{8}+{\left(\cos \left(\frac{7\pi }{6}\right)+i\sin \left(\frac{7\pi }{6}\right)\right)}^{2019}\right\}$

$=2^{2019}\left\{\left(\cos \left(2016\times \frac{11\pi }{6}\right)+i\sin \left(2016\times \frac{11\pi }{6}\right)\right)\frac{1}{8}+\left(\cos \left(2019\times \frac{7\pi }{6}\right)+i\sin \left(2019\times \frac{7\pi }{6}\right)\right)\right\}$

$=2^{2019}\left\{\left(\cos \left(2\pi \right)+i\sin \left(2\pi \right)\right)\frac{1}{8}+\left(\cos \left(\frac{3\pi }{2}\right)+i\sin \left(\frac{3\pi }{2}\right)\right)\right\}$

$=2^{2019}\left\{\frac{1}{8}+-i\right\}$

Hence, imaginary part is $-2^{2019}$