1 Week 3 3. Multivalued functions or dependences (continued) with the non-positive part of the real axis of the complex plane removed, and require Example 1: Consider the multivalued function (2) (1)

2 (a) Argue that (1)–(2) and the proposed branch cut describe a single-valued function – i.e., if tracing f(z) along any allowable contour, we return to the starting point with the same value of f(z). (b) Find f(–1 – 0i) and f(–1 + 0i). upper edge lower edge (observe the capital “ A ” in “ Arg ”). The single-valued functions (1)–(2) can be written as Comment:

3 (1) The proposed branch cut separates all available branches of the logarithm. Recall that the general form of these is where k = 0, ±1, ±2, ±3.... (2) Most multivalued functions (including all of the previous examples) are multivalued because they involve arg z (which is multivalued). Comments: We’ll need the following definitions: ۞ A set S  R n is called open if every point of S has a neighbourhood consisting entirely of points that belong to S.

4 ۞ A set S is called connected if any two points of S can be joined by a continuous curve, all points of which belong to S. ۞ An open and connected set is called a domain. ۞ The boundary of a domain S is the set of points such that any neighbourhood of these points includes some points from S and some that do not belong to S. Example 2: Give an example of a domain on the (x, y) plane and identify its boundary. In both cases, the answer should be given in the form of an (in)equality involving x and y. ۞ A contour is a ‘closed’ path, i.e. one that has not endpoints.

5 ۞ A simply connected domain S is a domain, such that any contour in it can be continuously shrunk into a point while remaining in S. Example 3: Give examples of a simply connected domain and a non-simply- connected domain in R 2. ۞ Two curves in a domain S are called topologically equivalent if their endpoints coincide and the curves can be continuously transformed one into another while remaining in S. Topologically equivalent contours satisfy the same definition without the requirement of coinciding endpoints. Example 4: Give example of a domain in the complex plane such that topologically non-equivalent contours exist in it.

6 on the complex plane with the segment [–1, 1] of the real axis removed, and require Example 5: Consider the dependence Do (3)–(4) and the proposed branch cut describe a single-valued function? (4) (3) If it does, what’s the value of f(0 + i0) ?

7 Comments: (1) To prove that a function f(z) is single-valued, it’s sufficient to show that the change of f(z) along all allowable contours is zero. (2) It’s sufficient to verify the zero change of only arg f, as | f | certainly restores its initial value. Consider two contours: one enclosing the cut and another, not enclosing it. The change of arg f along the latter is certainly zero, and the same applies to all contours that are topologically equivalent to the one under consideration. Example 5 (continued): The examine the change of arg f along the contour enclosing the cut, make up the following table...

8 z = 2 (if coming from z = –2, through the lower semi-plane) the point & from where you are coming arg (z + 1)arg (z – 1)arg f z = 2 z = –2 (if coming from z = –2, through the upper semi-plane) Tracing f(z) from z = 2 [where it’s given by (4)], one can deduce that f(0 + i0) = i.

9 and the condition Example 6: Consider the multivalued function Propose a branch cut making this function single-valued, and calculate f(0 + i0). The branch points in this example are the same as in Example 12 – hence, let’s first try the same branch cut.

10 z = 2 (if coming from z = –2, through the lower semi-plane) the point & from where you are coming arg (z + 1)arg (z – 1) 2 Im f z = 2 z = –2 (if coming from z = –2, through the upper semi-plane) Comment: Unlike the previous example (involving a root), with logarithms it’s more convenient to trace Im f, not arg f.

11 Example 6 (continued): Hence, this branch cut doesn’t work. Alternatively, consider removing from the real axis the (semi- infinite) segments [–∞, –1] and [1, +∞]. To check, whether this works, check the change of Im f(z) along some typical contours. Finally, it’s easy to calculate that f(0 + i0) = f(0) = iπ.