Convert the multivalued function $f(z) = \operatorname{ln} r + \mathrm{i} \theta \ (z = r \mathrm{e}^{\mathrm{i} \theta})$,
$$
0 < \theta < 5 \cdot \frac{\pi}{2}
$$
into a single-valued analytic branch.

Options:
A. $$
f_{1} ( z )=\operatorname{l n} r+\mathrm{i} \theta, D_{1} ; 0 < \theta< \pi;
$$
$$
f_{2} \left( z \right)=\operatorname{l n} r+\mathrm{i} \theta, D_{2}, \pi < \theta< 2 \pi ;
$$
$$
f_{3} \left( z \right)=\operatorname{l n} r+\mathrm{i} \theta, D_{3} ; 2 \pi< \theta< 3 \pi;
$$
$$
f_{4} \left( z \right)=\operatorname{l n} r+\mathrm{i} \theta, D_{4} : -\pi < \theta< 0.
$$
B. $$
f_{1} ( z )=\operatorname{l n} r+\mathrm{i} \theta, D_{1} ; -\pi< \theta< \pi;
$$
$$
f_{2} \left( z \right)=\operatorname{l n} r+\mathrm{i} \theta, D_{2}, 0 < \theta< \pi;
$$
$$
f_{3} \left( z \right)=\operatorname{l n} r+\mathrm{i} \theta, D_{3} ; \pi< \theta< 3 \pi;
$$
$$
f_{4} \left( z \right)=\operatorname{l n} r+\mathrm{i} \theta, D_{4} : 2 \pi< \theta< 4 \pi.
$$
C. $$
f_{1} ( z )=\operatorname{l n} r+\mathrm{i} \theta, D_{1} ; - \frac{\pi}{2} < \theta< \pi;
$$
$$
f_{2} \left( z \right)=\operatorname{l n} r+\mathrm{i} \theta, D_{2}, \pi< \theta< 2 \pi ;
$$
$$
f_{3} \left( z \right)=\operatorname{l n} r+\mathrm{i} \theta, D_{3} ; 2 \pi< \theta< 4 \pi;
$$
$$
f_{4} \left( z \right)=\operatorname{l n} r+\mathrm{i} \theta, D_{4} : \frac{5 \pi}{2} < \theta< 6 \pi.
$$
D. $$
f_{1} ( z )=\operatorname{l n} r+\mathrm{i} \theta, D_{1} ; -\pi< \theta< 0;
$$
$$
f_{2} \left( z \right)=\operatorname{l n} r+\mathrm{i} \theta, D_{2}, 0 < \theta< \pi ;
$$
$$
f_{3} \left( z \right)=\operatorname{l n} r+\mathrm{i} \theta, D_{3} ; \pi< \theta< 2 \pi;
$$
$$
f_{4} \left( z \right)=\operatorname{l n} r+\mathrm{i} \theta, D_{4} : 2 \pi< \theta< 3 \pi.
$$
E. $$
f_{1} ( z )=\operatorname{l n} r+\mathrm{i} \theta, D_{1} ; 0 < \theta< \frac{\pi}{2};
$$
$$
f_{2} \left( z \right)=\operatorname{l n} r+\mathrm{i} \theta, D_{2}, \frac{\pi} {2} < \theta< \pi ;
$$
$$
f_{3} \left( z \right)=\operatorname{l n} r+\mathrm{i} \theta, D_{3} ; \pi< \theta< 2 \pi;
$$
$$
f_{4} \left( z \right)=\operatorname{l n} r+\mathrm{i} \theta, D_{4} : 3 \pi < \theta< 5 \pi.
$$
F. $$
f_{1} ( z )=\operatorname{l n} r+\mathrm{i} \theta, D_{1} ; 0 < \theta< \pi/3;
$$
$$
f_{2} \left( z \right)=\operatorname{l n} r+\mathrm{i} \theta, D_{2}, \pi/3 < \theta< 2 \pi/3 ;
$$
$$
f_{3} \left( z \right)=\operatorname{l n} r+\mathrm{i} \theta, D_{3} ; 2 \pi/3 < \theta< 5 \pi/3;
$$
$$
f_{4} \left( z \right)=\operatorname{l n} r+\mathrm{i} \theta, D_{4} : 5 \pi/3 < \theta< 2 \pi.
$$
G. $$
f_{1} ( z )=\operatorname{l n} r+\mathrm{i} \theta, D_{1} ; 0 < \theta< 2 \pi;
$$
$$
f_{2} \left( z \right)=\operatorname{l n} r+\mathrm{i} \theta, D_{2}, 2 \pi< \theta< 3 \pi ;
$$
$$
f_{3} \left( z \right)=\operatorname{l n} r+\mathrm{i} \theta, D_{3} ; 3 \pi< \theta< 4 \pi;
$$
$$
f_{4} \left( z \right)=\operatorname{l n} r+\mathrm{i} \theta, D_{4} : - \pi < \theta< 0.
$$
H. $$
f_{1} ( z )=\operatorname{l n} r+\mathrm{i} \theta, D_{1} ; 0 < \theta< \pi;
$$
$$
f_{2} \left( z \right)=\operatorname{l n} r+\mathrm{i} \theta, D_{2}, \frac{\pi} {2} < \theta< \frac{3 \pi} {2} ;
$$
$$
f_{3} \left( z \right)=\operatorname{l n} r+\mathrm{i} \theta, D_{3} ; \pi< \theta< 2 \pi;
$$
$$
f_{4} \left( z \right)=\operatorname{l n} r+\mathrm{i} \theta, D_{4} : \frac{3 \pi} {2} < \theta< \frac{5 \pi} {2}.
$$
I. $$
f_{1} ( z )=\operatorname{l n} r+\mathrm{i} \theta, D_{1} ; 0 < \theta< \frac{\pi}{2};
$$
$$
f_{2} \left( z \right)=\operatorname{l n} r+\mathrm{i} \theta, D_{2}, \frac{\pi} {2} < \theta< \pi ;
$$
$$
f_{3} \left( z \right)=\operatorname{l n} r+\mathrm{i} \theta, D_{3} ; \pi< \theta< \frac{3 \pi} {2};
$$
$$
f_{4} \left( z \right)=\operatorname{l n} r+\mathrm{i} \theta, D_{4} : \frac{3 \pi} {2} < \theta< 2 \pi.
$$
Answer: H