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$$z=(-16i)^{5/4}=32(-i)^{5/4}$$ Use $$-i=cis\left(\frac{3\pi}{2}+2k\pi\right)$$ Then $$(-i)^{5/4}=(-i)^{1/4}=cis\left(\frac{3\pi}{8}+\frac{1}{2}k\pi\right)$$ choose $k \in \{0,1,2,3\}$. So the solutions will be: $$k=0 \rightarrow z_0=32\cdot cis\left(\frac{3\pi}{8}\right)\\ k=1 \rightarro...