14—Complex Variables 359
Asλvaries from 1 to∞, the two roots travel from− 1 → −∞and from− 1 → 0 , soz+stays inside
the unit circle (problem14.19). The integral is then
−
2 i
b
∫
C
dz
z^2 + 2λz+ 1
=−
2 i
b
∫
C
dz
(z−z+)(z−z−)
=−
2 i
b
2 πiRes
z=z+
=−
2 i
b
2 πi
1
z+−z−
=
2 π
b
√
λ^2 − 1
=
2 π
√
a^2 −b^2
14.7 Branch Points
Before looking at any more uses of the residue theorem, I have to return to the subject of branch
points. They are another type of singularity that an analytic function can have after poles and essential
singularities.
√
zprovides the prototype.
Thedefinitionof the word function, as in section12.1, requires that it be single-valued. The
function
√
zstubbornly refuses to conform to this. You can get around this in several ways: First,
ignore it. Second, change the definition of “function” to allow it to be multiple-valued. Third, change
the domain of the function.
You know I’m not going to ignore it. Changing the definition is not very fruitful. The third
method was pioneered by Riemann and is the right way to go.
The complex plane provides a geometric picture of complex numbers, but when you try to handle
square roots it becomes a hindrance. It isn’t adequate for the task. There are several ways to develop
the proper extension, and I’ll show a couple of them. The first is a sort of algebraic way, and the second
is a geometric interpretation of the first way. There are other, even more general methods, leading into
the theory of Riemann Surfaces and their topological structure, but I won’t go into those.
Pick a base point, sayz 0 = 1, from which to start. This will be a kind of fiduciary point near
which Iknowthe values of the function. Every other point needs to be referred to this base point. If I
state that the square root ofz 0 is one, then I haven’t run into trouble yet. Take another pointz=reiθ
and try to figure out the square root there.
√
z=
√
reiθ=
√
reiθ/^2 or
√
z=
√
rei(θ+2π)=
√
reiθ/^2 eiπ
The key question:How do you get fromz 0 toz?What path did you select?
0 1 2 −^3
z z z z
. In the picture,zappears to be at about 1. 5 e^0.^6 ior so.
. On the path labeled 0, the angleθstarts at zero atz 0 and increases to 0.6 radians, so√reiθ/^2 varies
continuously from 1 to about 1. 25 e^0.^3 i.
. On the path labeled 1, angleθagain starts at zero and increases to 0 .6 + 2π, so√reiθ/^2 varies
continuously from 1 to about 1. 25 e(π+0.3)i, which is minus the result along path #0.
. On the path labeled 2, angleθgoes from zero to 0 .6+4π, and√reiθ/^2 varies from 1 to 1. 25 e(2π+0.3)i
and that is back to the same value as path #0.
. For the path labeled− 3 , the angle is 0. 6 − 6 π, resulting in the same value as path #1.
