# Do we need a branch cut to choose a branch?

If we arbitarily choose the initial value $$w=\ln(2 \sqrt{2}) + i \frac{\pi}{4}$$ for $$\ln(2 +2i)$$ , then as $$z$$ travels along a loop the encircles the origin $$\nu$$ times, $$\log(z)$$ moves along a path from $$\omega$$ to $$\omega + 2 \nu \pi i$$. Check that you understand (roughly the shapes of the illustrated paths

Pg-98,99, Tristan Needham Visual Complex Analysis

1. How can he speak of a multi-valued function as a single valued one without a branch-cut?

2. If chooses a branch for evaluating the multi-valued function as a single valued function, would that implicitly put a branch cut? I think it does, and then, I am confused because how he was able to loop back along the dark circle back to $$2 + 2i$$ because the points on the positive $$x$$-axis are removed.

Because if he would have done a branch cut while choosing a branch, then why can he move around this curve in the picture, I mean then there would be a branch cut so a "line" where he can't go further since otherwise he would get to the multivalued issue again.

Obviously $$\log(z)$$ is a multivalued function as explained on p.98. To define single-valued branches of $$\log(z)$$ the author introduces a branch cut from $$0$$ to $$\infty$$ and says that the most popular choice is the negative real axis (see p.99). This should answer your first question: The author does not speak of a multi-valued function as a single-valued one without a branch-cut.

After having made the branch cut we can define the principal branch of the logarithm based on the principal value of $$\operatorname{Arg}(z)$$. This is a function

$$\operatorname{Log} :\mathbb C \setminus \{0\} \to \mathbb C .$$

The problem is that $$\operatorname{Log}$$ is not continuous at the points of the branch cut. You can still move along any closed curve $$u : [0,1] \to \mathbb C \setminus \{0\}$$, and always end with the same value of $$\operatorname{Log}(z)$$. But this is not a valuable information, $$\operatorname{Log} \circ \ u : [0,1] \to \mathbb C$$ is not continuous if $$u$$ crosses the branch cut.

Unfortunately Needham does not say that he wants to get continuous branches. Anyway, one can easily show that there does not exist a continuous $$\operatorname{Log} :\mathbb C \setminus \{0\} \to \mathbb C$$. This can be resolved by removing the branch cut $$B$$ which gives a continuous $$\operatorname{Log} :\mathbb C \setminus B \to \mathbb C .$$ In this situation he is in fact no longer able to loop back along the two circles in Figure [36] back to $$2+2i$$ because the points of the negative $$x$$-axis are removed. But be aware that his original $$\operatorname{Log}$$ is defined on $$\mathbb C \setminus \{0\}$$.

To define a "single-valued branch" of the logarithm without any further assumpions on $$\operatorname{Log}$$ we can choose for each $$z \in \mathbb C \setminus \{0\}$$ an arbitrary element $$\operatorname{Log}(z) \in e^{-1}(z)$$. The resulting function will in general be highly discontinuous. The principal branch is a special choice which is as continuous as possible; the discontinuities lie on the branch cut.

• First of all thank you for your answer and sorry that I respond so late. Did I understand it correctly that in his example the function is not continuous? So I mean only when we remove a branch cut we get a continuous single valued function which we can work with. But removing this branch cut also leads to the "problem" that we can't turn around the origin. Jun 13 at 11:23
• @Wave Yes, this is correct. Jun 13 at 12:30
• Ah but then even if it's not continuous he can move along it. Jun 13 at 12:57
• But for computations we always need to remove a branch cut right? Jun 13 at 12:57
• @Wave It depends on what you want to do. If you just want to assoiciate to $z$ a single value of $\log z$ and do not care about continuity, then you do not need to exclude the branch cut. But if continuity is essential in some context, you must remove it. As a branch cut we can choose any ray starting at $0$ an going to $\infty$. At the points of the branch cut the $arg$-function jumps by a value of $2\pi$ if you travel along a circle. Jun 13 at 13:04