Argument Principle

Let {z_0} be a zero of a meromorphic {f} with multiplicity {m}. Then we can write {f(z)=(z-z_0)^m\cdot g(z)} where {g(z_0)\ne0}. Taking derivatives yields

\displaystyle f'(z)=m(z-z_0)^{m-1}\cdot g(z)+(z-z_0)^m\cdot g'(z).

Hence, for {f'/f}, we get

\displaystyle \displaystyle\frac{f'(z)}{f(z)}=\frac{m}{z-z_0}+\frac{g'(z)}{g(z)}.

The residue of this sum is simply the sum of the residues of the two parts. The first term has residue {m} at {z_0}. The second part has no pole at {z_0}, and hence zero residue. Thus, the residue of {f'/f} at {z_0} is {m}.

What if we pick a pole of {z_p} of {f}? Then by a similar construction, if {z_p} is an order {q} pole, we can write {f(z)=(z-z_p)^{-q}\cdot h(z)} and compute

\displaystyle \displaystyle\frac{f'(z)}{f(z)}=\frac{-q}{z-z_p}+\frac{h'(z)}{h(z)},

yielding a total residue of {-q}.

If a point {z} is neither a pole nor a zero, then {f'/f} is holomorphic at {z}, and has residue zero at {z}.

If we take a big contour around all of these points, then the integral will be the sum of the residues inside the contour, which we have just shown is the number of roots minus the number of poles (counted with multiplicity and order, respectively).

That is, if {N_\gamma(f)} denotes the number of zeros counted with multiplicity inside a contour {\gamma}, and {P_\gamma(f)} denotes the number of poles counted with order, then

\displaystyle \displaystyle\frac{1}{2\pi i}\int_\gamma \frac{f'(z)}{f(z)}dz=N_\gamma(f)-P_\gamma(f)

This result is known as the argument principle.

2 Responses to Argument Principle

  1. Winston says:

    Awesome stuff. Complex analysis is my favorite. Thanks for writing this. It’s quite refreshing looking back on all this again.

    Also, just an fyi: you probably mean m and not k in that first equation you have there.


Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s