Mathematical Tools for Physics

14—Complex Variables 442

Whichvalue ofln(−a)do I take? That answer is dictated by how I arrived at the point−awhen I pushed the
contour fromC 1 toC 2. That is,lna+iπ.

−lnz

z

(a+z)^3

=−

[

lna+iπ−

1

a

(z+a)−

1

a^2

(z+a)^2

2

+···

]

[

(z+a)−a

] 1

(z+a)^3

I’m interested only in the residue, so I want only the coefficient of the power 1 /(z+a). That is

−

[

−

1

a

−

1

2 a^2

(−a)

]

=

1

2 a

Did I have to do all this work to get this answer? Absolutely not. This falls under the classic heading of using a
sledgehammer as a fly swatter. It does show the technique though, and in the process I had an excuse to show
that third-order poles needn’t be all that intimidating.

14.9 Other Results
Polynomials: There are some other consequences of looking in the complex plane that are very different from
any of the preceding. If you did problem3.11, you realize that the functionez= 0has no solutions, even in the
complex plane. You’re used to finding roots of equations such as quadratics and maybe you’ve even encountered
the cubic formula too. How do you know that every polynomial even has a root? Maybe there’s an order-137
polynomial that has none. No, it doesn’t happen. That every polynomial has a root (nof them in fact) is
the Fundamental Theorem of Algebra. Gauss proved it first, but after the advent of complex variable theory it
becomes an elementary exercise.
A polynomial isf(z) =anzn+an− 1 zn−^1 +···+a 0. Consider the integral
∫

C

dz

f′(z)

f(z)

around a large circle.f′(z) =nanzn−^1 +···, so this is

∫

C

dz

nanzn−^1 + (n−1)an− 1 zn−^2 +···

anzn+an− 1 zn−^1 +···+a 0

=

∫

C

dz

n

z

1 +(n−na1)nazn−^1 +···

1 +aann−z^1 +···