Multivalued functions and branch cuts

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Multivalued functions and branch cuts
On the RHS let us write t as follows:
t = r exp[i(θ + 2kπ)],
where k is an integer. We then obtain
(θ + 2kπ)
ln r + i
(θ + 2kπ)
= r exp i
t1/n = exp
where k = 0, 1, . . . , n − 1; for other values of k we simply recover the roots already found.
Thus t has n distinct nth roots. 24.5 Multivalued functions and branch cuts
In the definition of an analytic function, one of the conditions imposed was
that the function is single-valued. However, as shown in the previous section, the
logarithmic function, a complex power and a complex root are all multivalued.
Nevertheless, it happens that the properties of analytic functions can still be
applied to these and other multivalued functions of a complex variable provided
that suitable care is taken. This care amounts to identifying the branch points of
the multivalued function f(z) in question. If z is varied in such a way that its
path in the Argand diagram forms a closed curve that encloses a branch point,
then, in general, f(z) will not return to its original value.
For definiteness let us consider the multivalued function f(z) = z 1/2 and express
z as z = r exp iθ. From figure 24.1(a), it is clear that, as the point z traverses any
closed contour C that does not enclose the origin, θ will return to its original
value after one complete circuit. However, for any closed contour C that does
enclose the origin, after one circuit θ → θ + 2π (see figure 24.1(b)). Thus, for the
function f(z) = z 1/2 , after one circuit
r 1/2 exp(iθ/2) → r 1/2 exp[i(θ + 2π)/2] = −r 1/2 exp(iθ/2).
In other words, the value of f(z) changes around any closed loop enclosing the
origin; in this case f(z) → −f(z). Thus z = 0 is a branch point of the function
f(z) = z 1/2 .
We note in this case that if any closed contour enclosing the origin is traversed
twice then f(z) = z 1/2 returns to its original value. The number of loops around
a branch point required for any given function f(z) to return to its original value
depends on the function in question, and for some functions (e.g. Ln z, which also
has a branch point at the origin) the original value is never recovered.
In order that f(z) may be treated as single-valued, we may define a branch cut
in the Argand diagram. A branch cut is a line (or curve) in the complex plane
and may be regarded as an artificial barrier that we must not cross. Branch cuts
are positioned in such a way that we are prevented from making a complete
Figure 24.1 (a) A closed contour not enclosing the origin; (b) a closed
contour enclosing the origin; (c) a possible branch cut for f(z) = z 1/2 .
circuit around any one branch point, and so the function in question remains
For the function f(z) = z 1/2 , we may take as a branch cut any curve starting
at the origin z = 0 and extending out to |z| = ∞ in any direction, since all
such curves would equally well prevent us from making a closed loop around
the branch point at the origin. It is usual, however, to take the cut along either
the real or the imaginary axis. For example, in figure 24.1(c), we take the cut as
the positive real axis. By agreeing not to cross this cut, we restrict θ to lie in the
range 0 ≤ θ < 2π, and so keep f(z) single-valued.
These ideas are easily extended to functions with more than one branch point.
Find the branch points of f(z) =
branch cuts.
We begin by writing f(z) as
f(z) =
z 2 + 1, and hence sketch suitable arrangements of
z2 + 1 =
(z − i)(z + i).
As shown above, the function g(z) = z 1/2 has a branch point at z = 0. Thus we might
expect f(z) to have branch points at values of z that make the expression under the square
root equal to zero, i.e. at z = i and z = −i.
As shown in figure 24.2(a), we use the notation
z − i = r1 exp iθ1
z + i = r2 exp iθ2 .
We can therefore write f(z) as
f(z) = r1 r2 exp(iθ1 /2) exp(iθ2 /2) = r1 r2 exp i(θ1 + θ2 )/2 .
Let us now consider how f(z) changes as we make one complete circuit around various
closed loops C in the Argand diagram. If C encloses
(i) neither branch point, then θ1 → θ1 , θ2 → θ2 and so f(z) → f(z);
(ii) z = i but not z = −i, then θ1 → θ1 + 2π, θ2 → θ2 and so f(z) → −f(z);
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