Question:

Step 1: Apply the Angle Bisector Theorem for the internal bisector. Using the Angle Bisector Theorem, the internal bisector of $\angle APB$ divides the line segment $AB$ internally in the ratio $AP:PB$. Given that this ratio is $3:1$, the internal bisector meets the line $AB$ at a point $P_1$ such that $AP_1:P_1B = AP:PB = 3:1$. Step 2: Apply the Angle Bisector Theorem for the external bisector. Similarly, the external bisector of $\angle APB$ divides the line segment $AB$ externally in the same ratio $AP:PB = 3:1$. This external bisector meets the line $AB$ at a point $P_2$ such that $AP_2:P_2B = AP:PB = 3:1$. Step 3: Establish the perpendicularity of the angle bisectors. The internal and external angle bisectors of any angle are always perpendicular to each other. In this case, $PP_1$ is the internal bisector and $PP_2$ is the external bisector of $\angle APB$. Therefore, the angle formed by these two bisectors at point $P$ is a right angle, i.e., $\angle P_1PP_2 = \frac{\pi}{2}$. Step 4: Determine the locus of point P. Since $\angle P_1PP_2 = \frac{\pi}{2}$, point $P$ subtends a right angle at the diameter $P_1P_2$. This implies that $P$ must lie on a circle for which $P_1P_2$ is a diameter. Step 5: State the position of point B relative to the circle. The problem statement also indicates that point $B(z_2)$ lies inside this circle. This is an additional condition or observation about the specific setup of the problem.