4. In the complex plane, the maximum internal angle of the triangle formed by the points corresponding to the complex numbers $\frac{1}{\mathrm{i}}$, $\frac{2}{\mathrm{i}-1}$, and $(\mathrm{i}-1)^{3}$ is equal to ( ).
(A) $\pi-\arccos \frac{2}{\sqrt{13}}$
(B) $-\arccos \frac{2}{\sqrt{13}}$
(C) $45^{\circ}$
(D) $120^{\circ}$
$$
\begin{array}{l}
\text { 4. A. } \\
\text { Given } \frac{1}{i}=-i, \frac{2}{i-1}= \\
-1-i, (i-1)^{3}=2+2 i
\end{array}
$$

we know that the points corresponding to them are $A$, 
$B$, and $C$ as shown in Figure 2. The largest interior angle of $\triangle ABC$ is $\angle A$.
$$
\begin{aligned}
& \text { Given } \arccos A \\
= & \arccos \frac{A B^{2}+A C^{2}-B C^{2}}{2|A B \cdot A C|} \\
= & \arccos \frac{1+13-18}{2 \sqrt{13}}=\arccos \left(-\frac{2}{\sqrt{13}}\right) .
\end{aligned}
$$

Thus, $\angle A=\pi-\arccos \frac{2}{\sqrt{13}}$.