Lecture 23: the argument principle

Let $f:D\to ℂ$ be an analytic function, and suppose we want to study the zeros of $f$. One method would be to study integrals of $\frac{1}{f}$ along some closed contour; then the residue theorem in a sense “counts” the poles of $\frac{1}{f}$ inside this contour, and so “counts” the zeros of $f$. However, this counting is really adding up the residues, which is pretty different: for example, it’s easy for the residues at different poles to cancel out.

A different way to turn zeros into poles is to take the logarithm. However, the logarithm of even a simple zero is actually an essential singularity, so this isn’t ideal; and of course the logarithm needs a branch cut to be analytic. We could solve both of these problems by differentiating: while e.g. $\log (z-z_0)$ is not as well-behaved as we would like, $\frac{d}{dz}\log (z-z_0)=\frac{1}{z-z_0}$ is as well-behaved as we could ask, for a pole. Does this avoid the residue issue above?

More generally, if $f$ is analytic on $D$ and has a zero at $z_0$, the analogue of the above is $\frac{d}{dz}\log f(z)=\frac{f^{\prime }(z)}{f(z)}$. Since $f$ and $f^{\prime }$ are analytic, we can try to evaluate the residue of $\frac{f^{\prime }}{f}$ at $z_0$. For example, if $f$ has a simple zero at $z_0$, then $\frac{f^{\prime }}{f}$ has residue $\frac{f^{\prime }(z_0)}{f^{\prime }(z_0)}=1$! So the logarithmic derivative turns simple zeros into simple poles of residue $1$, so the sum of the residues at these poles really does give a count of the zeros.

What about higher order zeros? If $f$ has a zero of order $n$ at $z_0$, i.e. $f(z)=g(z){(z-z_0)}^n$ for some analytic function $g(z)$ with $g(z_0)\ne 0$, then

$$\frac{f^{\prime }(z)}{f(z)}=\frac{g^{\prime }(z){(z-z_0)}^n+ng(z){(z-z_0)}^{n-1}}{g(z){(z-z_0)}^n}=\frac{g^{\prime }(z)}{g(z)}+\frac{n}{z-z_0}.$$

Since $g(z_0)\ne 0$, the first term is analytic at $z_0$, and so the residue of $\frac{f^{\prime }}{f}$ at $z_0$ is equal to the residue of $\frac{n}{z-z_0}$, i.e. $n$. Therefore summing up the residues of $\frac{f^{\prime }}{f}$ is equivalent to counting the zeros of $f$, with multiplicity.

We could also allow $f$ to be meromorphic, rather than holomorphic, on $D$, i.e. have poles as well as zeros. Viewing a pole of order $n$ as a zero of order $-n$, the same argument above implies that if $f$ has a pole of order $n$ at $z_0$, then $\frac{f^{\prime }}{f}$ has a simple pole at $z_0$ with residue $-n$. Therefore for a meromorphic function $f$ we can think of summing up the residues of $\frac{f}{f^{\prime }}$ as the number of zeros of $f$, with multiplicity, minus the number of poles, with multiplicity. In other words, via the residue theorem, we have proven the following theorem:

Theorem (Argument principle).

Let $D$ be a bounded domain with piecewise smooth boundary and $f$ a meromorphic function on $D$ which extends to an analytic function on $\partial D$, with $f(z)\ne 0$ for $z\in \partial D$. Let $N_0$ be the number of zeros of $f$ in $D$, counted with multiplicity, and $N_{\mathrm{\infty }}$ the number of poles, also counted with multiplicity. Then

$$\frac{1}{2\pi i}{\int }_{\partial D}\frac{f^{\prime }(z)}{f(z)}𝑑z=N_0-N_{\mathrm{\infty }}.$$

Indeed, by the residue theorem the left-hand side is the sum of the residues of $\frac{f^{\prime }}{f}$, which gives the right-hand side by the discussion above.

More generally, for a closed path $\gamma$ in $D$ along which $f$ is analytic and nonzero, we call

$$\frac{1}{2\pi i}{\int }_{\gamma }\frac{f^{\prime }(z)}{f(z)}𝑑z$$

the logarithmic integral of $f$ along $\gamma$. Since the integrand is the logarithmic derivative, we could rewrite it as

$$\frac{1}{2\pi i}{\int }_{\gamma }d\log f(z).$$

Note however that $\log f(z)$ need not be analytic on $D$, hence we cannot simply evaluate this by the fundamental theorem of calculus (or else it would be zero for $\gamma$ closed); $d\log f(z)$ is a closed differential, but is not exact in general.

To justify the terminology, recall that $\log z=\log |z|+i\arg z$, so $\log f(z)=\log |f(z)|+i\arg f(z)$. Therefore we can rewrite the above as

$$\frac{1}{2\pi i}{\int }_{\gamma }d\log |f(z)|+\frac{1}{2\pi }{\int }_{\gamma }d\arg f(z).$$

The first differential $d\log |f(z)|$, despite the non-analyticity of $\log |f(z)|$, is exact, and so the integral does not depend on the path, only on the endpoints of $\gamma$: if $\gamma :[a,b]\to ℂ$, then this is $\frac{1}{2\pi i}(\log |f(\gamma (b))|-\log |f(\gamma (a))|)$, so in particular for a closed path this is $0$. So we are left with the argument integral

$$\frac{1}{2\pi }{\int }_{\gamma }d\arg f(z),$$

which is not exact: $\arg f(z)$ is multivalued, and it may well happen that as we go around $\gamma$, the ending value of $\arg f(z)$ is different from the starting value, even though these are at the same point. Consider for example $f(z)=z$ and $\gamma$ the unit circle.

However, if we fix a continuous single-valued branch $A(t)$ of $\arg f(\gamma (t))$, then this integral is simply given by $\frac{1}{2\pi }(A(b)-A(a))$. Since two different choices of $A(t)$ differ by a constant, this is well-defined; up to the factor of $\frac{1}{2\pi }$, it is called the increase in argument of $f$ along $\gamma$. (So for $f(z)=z$ and $\gamma$ the unit circle as above, this would be $2\pi$, hence the integral would be $1$, compatibly with the argument principle.) By splitting the curve $\gamma$ into pieces and adding up the increases in argument of $f$ along each, we can often evaluate this integral, and hence apply the argument principle, without too much direct calculation.

To illustrate the theorem, consider the polynomial $p(z)=z^6+9z^4+z^3+2z+4$. How many zeros does $p$ have (counting multiplicity) in the first quadrant ?

Note first that $p$ has no zeros on the positive real line, since for $x\ge 0$ we have $p(x)\ge 4$. On the positive imaginary line, if $z=ix$ for $x\ge 0$ we have $p(ix)=-x^6+9x^4-i{x}^3+2ix+4$, so with real part $-x^6+9x^4+4$ and imaginary part $-x^3+2x$. The imaginary part vanishes at $x=0$ or $x=\pm \sqrt{2}$, and one can check that at both of these points the real part is nonzero, so $p(xi)\ne 0$ for all $x\ge 0$ as well.

Let $D$ be the region , ; so we want to count the limit of the number of zeros of $p$ contained in $D$ as $R\to \mathrm{\infty }$. We know that $p$ has no zeros along either of the straight edges contained in $\partial D$; the remaining part of the boundary is the arc from $R$ to $Ri$ of angle $\pi /2$ and radius $R$, and since $p$ has finitely many zeros there are no zeros on this arc once we choose $R$ large enough. Therefore we can apply the argument principle: since $p$ is holomorphic on $D$,

$$\frac{1}{2\pi i}{\int }_{\partial D}\frac{p^{\prime }(z)}{p(z)}𝑑z=\frac{1}{2\pi i}{\int }_{\partial D}d\log p(z)$$

is the number of zeros of $p$ in $D$.

For $z$ positive real, $p(z)$ is positive real, so $\arg p(z)=0$ and there is no increase in the argument as $z$ moves from $0$ to $R$ along the real axis. Along the quarter-circle of radius $R$, for $|z|=R$ large enough $p(z)\approx z^6$, so $\arg p(z)\approx \arg (z^6)=6\arg z$, so the increase in argument is approximately $6\cdot \frac{\pi }{2}=3\pi$. (This approximation is good enough because after dividing by $2\pi$, the integral should give an integer, so we can just take the nearest integer multiple of $2\pi$ at the end.) Finally, along the imaginary axis as $z$ goes from $Ri$ to $0$, we saw above that $p(ix)=-x^6+9x^4+4+(-x^3+2x)i$ for $x>0$ has imaginary part zero only at $\sqrt{2}$; for $x>\sqrt{2}$ it is negative, and for it is positive. At $Ri$, $p(Ri)\approx {(Ri)}^6=-R^6$, with argument approximately $\pi$, and the above shows that as $x$ goes from $R$ to $\sqrt{2}$, $p(xi)$ remains in the lower half-plane, and $p(i\sqrt{2})=-2^3+9\cdot 2^2+4=32$ with argument $2\pi$, so the increase in argument along this line segment is $\pi$. Finally, as $x$ goes from $\sqrt{2}$ to $0$, at both endpoints $p(z)$ is positive real, and in between it remains in the upper half-plane, so the increase in argument is $0$.

Therefore we have found that the total increase in argument is (in the limit as $R\to \mathrm{\infty }$) $0+3\pi +\pi +0=4\pi$, and so the number of zeros in the first quadrant is $\frac{4\pi }{2\pi }=2$. With the help of a computer, one could compute that these are at approximately $z\approx 0.0426+3.0087i$ and $z\approx 0.56725+0.64665i$.

A more conceptual application is Rouché’s theorem.

Theorem.

Let $D$ be a bounded domain with piecewise smooth boundary, and let $f,h$ be analytic functions on $D\cup \partial D$ such that for $z\in \partial D$. Then $f$ and $f+h$ have the same number of zeros in $D$ (counted with multiplicity).

In other words, we can perturb $f$ by a function which is “small” relative to $f$ in a certain sense without changing the number of zeros.

Proof.

Note that the assumptions imply $f(z)\ne 0$ for $z\in \partial D$. We have $f(z)+h(z)=f(z)\left(1+\frac{h(z)}{f(z)}\right)$, and since the second factor is in a disk of radius $1$ centered at $1$, and so in particular is in the right half-plane. Therefore $\arg (f(z)+h(z))=\arg f(z)+\arg \left(1+\frac{h(z)}{f(z)}\right)$, and the increase in argument of the second term as $z$ moves along $\partial D$ is $0$. Therefore the total increase in argument of $f+h$ is the same as that of $f$, so the claim follows from the argument principle. ∎

A fun consequence is the following slick proof of the fundamental theorem of algebra. Let $p(z)=a_n{z}^n+a_{n-1}{z}^{n-1}+\mathrm{\cdots }+a_{1}z+a_0$. On the boundary of a disk of sufficiently large radius $R$ centered at the origin, we have $|a_n{z}^n|>|a_{n-1}{z}^{n-1}+\mathrm{\cdots }+a_0|$, so $f$ has the same number of zeros within this disk as $a_n{z}^n$, which of course has $n$ zeros with multiplicity, namely a single zero of order $n$ at $z=0$.

A more concrete application is the following. Consider the equation $e^z=1+2z$. What are the solutions with ?

One solution is given by $z=0$. Are there any others?

Rewriting this as $e^z-1-2z=0$, we want to write the left-hand side as $f+h$ for some $f$ and $h$ satisfying the hypotheses of Rouché’s theorem: that is, on the unit circle, . Choosing $f(z)=-2z$ and $h(z)=e^z-1$ works: for $|z|=1$, we have $|-2z|=2$ and . So $e^z-1-2z$ has the same number of zeros in the unit disk as $-2z$, which has only a single zero at $z=0$, so we can answer the question above in the negative: $z=0$ is the only solution of $e^z=1+2z$ in the unit disk. (It is not however the only solution in $ℂ$, e.g. $z\approx 1.25643$ also works; in fact there are infinitely many solutions.)