5. B. Reason: Since $\arg (2+2 \mathrm{i})=\frac{\pi}{4}$, we have $\arg (2+2 \mathrm{i})^{4}=\pi, \arg (1-\sqrt{3} \mathrm{i})=\frac{5 \pi}{3}$, thus $\arg (1-\sqrt{3} \mathrm{i})^{5}=\frac{\pi}{3}$, hence $\arg \left[\frac{(2+2 \mathrm{i})^{4}}{(1-\sqrt{3} \mathrm{i})^{5}}\right]=\frac{2 \pi}{3}$, so the point corresponding to the complex number $\frac{(2+2 \mathrm{i})^{4}}{(1-\sqrt{3} \mathrm{i})^{5}}$ is in the second quadrant, therefore the answer is B.
5. (1996 National College Entrance Examination) The complex number $\frac{(2+2 \mathrm{i})^{4}}{(1-\sqrt{3} \mathrm{i})^{5}}$ equals ( ).
A. $1-\sqrt{3} \mathrm{i}$
B. $-1+\sqrt{3} i$
C. $1+\sqrt{3} \mathrm{i}$
D. $-1-\sqrt{3} \mathrm{i}$