# Onward to the argument principle and Rouche’s Theorem

Consider the function $f(z) = z^3$ and look at what it does to values on the circle $z(t) = e^{it}, t \in [0, 2\pi)$. $f(e^{it}) = e^{i3t}$ and as we go around the circle once, we see that $arg(f(z)) = arg(e^{i3t})$ ranges from $0$ to $6 \pi$. Note also that $f$ has a zero of order 3 at the origin and $(2 \pi)3 = 6 \pi$ That is NOT a coincidence.

Now consider the function $g(z) = \frac{1}{z^3}$ and look at what it does to values on the circle $z(t) = e^{it}, t \in [0, 2\pi)$. $g(e^{it}) = e^{-i3t}$ and as we go around the circle once, we see that $arg(g(z)) = arg(e^{-i3t})$ ranges from $0$ to $-6 \pi$. And here, $g$ has a pole of order 3 at the origin. This too, is not a coincidence.

We can formalize this somewhat: in the first case, suppose we let $\gamma$ be the unit circle taken once around in the standard direction and let’s calculate:

$\int_{\gamma} \frac{f'(z)}{f(z)} dz = \int_{\gamma}\frac{3z^2}{z^3}dz = 3 \int_{\gamma} \frac{1}{z}dz = 6 \pi i$

In the second case: $\int_{\gamma} \frac{g'(z)}{g(z)} dz = \int_{\gamma}\frac{-3z^{-4}}{z^{-3}}dz = -3 \int_{\gamma} \frac{1}{z}dz = -6 \pi i$

Here is what is going on: you might have been tempted to think $\int_{\gamma} \frac{f'(z)}{f(z)} dz = Log(f(z))|_{\gamma} = (ln|f(z)| +iArg(f(z)) )|_{\gamma}$ and this almost works; remember that $Arg(z)$ switches values abruptly along a ray from the origin that follows the negative real axis..and any version of the argument function must do so from SOME ray from the origin. The real part of the integral (the $ln(|f(z)|$ part) cancels out when one goes around the curve. The argument part (the imaginary part) does not; in fact we pick up a value of $2 \pi i$ every time we cross that given ray from the origin and in the case of $f(z) = z^3$ we cross that ray 3 times.

This is the argument principle in action.

Now of course, this principle can work in the vicinity of any isolated singularity or zero or along a curve that avoids singularities and zeros but encloses a finite number of them. The mathematical machinery we develop will help us with this concept.

So, let’s suppose that $f$ has a zero of order $m$ at $z = z_0$. This means that $f(z) = (z-z_0)^m g(z)$ where $g(z_0) \neq 0$ and $g$ is analytic on some open disk about $z_0$.

Now calculate: $\frac{f'(z)}{f(z)} dz = \frac{m(z-z_0)^{m-1} g(z) - (z-z_0)^m g'(z)}{(z-z_0)^m g(z)} = \frac{m}{z-z_0} + \frac{g'(z)}{g(z)}$. Now note that the second term of the sum is an analytic function; hence the Laurent series for $\frac{f'(z)}{f(z)}$ has $\frac{m}{z-z_0}$ as its principal part; hence $Res(\frac{f'(z)}{f(z)}, z_0) = m$

Now suppose that $f$ has a pole of order $m$ at $z_0$. Then $f(z) =\frac{1}{h(z)}$ where $h(z)$ has a zero of order $m$. So as before write $f(z) = \frac{1}{(z-z_0)^m g(z)} = (z-z_0)^{-m}(g(z))^{-1}$ where $g$ is analytic and $g(z_0) \neq 0$. Now $f'(z) = -m(z-z_0)^{-m-1}g(z) -(g(z))^{-2}g'(z)(z-z_0)^{-m}$ and
$\frac{f'(z)}{f(z)} =\frac{-m}{z-z_0} - \frac{g'(z)}{g(z)}$ where the second term is an analytic function. So $Res(\frac{f'(z)}{f(z)}, z_0) = -m$

This leads to the following result: let $f$ be analytic on some open set containing a piecewise smooth simple closed curve $\gamma$ and analytic on the region bounded by the curve as well, except for a finite number of poles. Also suppose that $f$ has no zeros on the curve.

Then $\int_{\gamma} \frac{f'(z)}{f(z)} dz = 2 \pi i (\sum^k_{j =1} m_k - \sum^l_{j = 1}n_j )$ where $m_1, m_2...m_k$ are the orders of the $k$ zeros of $f$ inside of $\gamma$ and $n_1, n_2.., n_l$ are the orders of the poles of $f$ inside $\gamma$.

This follows directly from the theory of cuts:

Use of our result: let $f(z) = \frac{(z-i)^4(z+2i)^3}{z^2 (z+3i-4)}$ and let $\Gamma$ be a circle of radius 10 (large enough to enclose all poles and zeros of $f$. Then $\int_{\Gamma} \frac{f'(z)}{f(z)} dz = 2 \pi i (4 + 3 -2-1) = 8 \pi i$. Now if $\gamma$ is a circle $|z| = 3$ we see that $\gamma$ encloses the zeros at $i, -2i,$ and the pole at 0 but not the pole at $4-3i$ so $\int_{\Gamma} \frac{f'(z)}{f(z)} dz = 2 \pi i (4+3 -2) = 10 \pi i$

Now this NOT the main use of this result; the main use is to describe the argument principle and to get to Rouche’s Theorem which, in turn, can be used to deduce facts about the zeros of an analytic function.

Argument principle: our discussion about integrating $\frac{f'(z)}{f(z)}$ around a closed curve (assuming that the said curve runs through no zeros of $f$ and encloses a finite number of poles ) shows that, as we traverse the curve, the argument of the function changes by $2 \pi (\text{ no. of zeros - no. of poles})$ where the zeros and poles are counted with multiplicities.

Example: consider the function $f(z) = z^8 + z^2 + 1$. Let’s find how many zeros it has in the first quadrant.

If we consider the quarter circle of very large radius $R$ (that stays in the first quadrant and is large enough to enclose all first quadrant zeros) and note $f(Re^{it}) = R^8e^{i8t}(1+ \frac{1}{R^6}e^{-i6t} + \frac{1}{R^8 e^{i8t}})$ we see that the argument changes by about $8(\frac{\pi}{2} = 4 \pi$. The function has no roots along the positive real axis. Now setting $z = iy$ to run along the positive imaginary axis we get $f(iy) = y^8 -y^2 + 1$ which is positive for large $R$, has one relative minimum at $2^{\frac{-1}{3}}$ which yields a positive number, and is zero at $z = 0$. So the argument stays at $4 \pi$ so, we get $4 \pi = 2 \pi (\text{no. of roots in the first quadrant})$ which means that we have 2 roots in the first quadrant.

In fact, you can find an online calculator which estimates them here.

Now for Rouche’s Theorem
Here is Rouche’s Theorem: let $f, g$ be analytic on some piecewise smooth closed curve $C$ and on the region that $C$ encloses. Suppose that, on $C$ we have $|f(z) + g(z)| < |f(z)|$. Then $f, g$ have the same number of zeros inside $C$. Note: the inequality precludes $f$ from having a zero on $C$ and we can assume that $f, g$ have no common zeros, for if they do, we can “cancel them out” by, say, writing $f(z) = (z-z_0)^m f_1(z), g(z) = (z-z_0)^mg_1(z)$ at the common zeros. So now, on $C$ we have $|1 + \frac{g(z)}{f(z)}| < |1|$ which means that the values of the new function $\frac{g(z)}{f(z)}$ lie within the circle $|w+1| < 1$ in the domain space. This means that the argument of $\frac{g(z)}{f(z)}$ has to always lie between $\frac{\pi}{2}$ and $\frac{3 \pi }{2}$ This means that the argument cannot change by $2 \pi$ so, up to multiplicity, the number of zeros and poles of $\frac{g(z)}{f(z)}$ must be equal. But the poles come from the zeros of the denominator and the zeros come from the zeros of the numerator.

And note: once again, what happens on the boundary of a region (the region bounded by the closed curve) determines what happens INSIDE the curve.

Now let’s what we can do with this. Consider our $g(z) = z^8 + z^2 + 1$. Now $|z^8 -(z^8 + z^2 + 1)| =|z^2+1| < |z^8|$ for $R = \frac{3}{2}$ (and actually smaller). This means that $z^8$ and $z^8+z^2 + 1$ have the same number of roots inside the circle $|z| = \frac{3}{2}$: eight roots (counting multiplicity). Now note that $|z^8 +z^2 + 1 -1| = |z^8+z^2| < |1|$ for $|z| \leq \frac{2}{3}$ So $z^8 +z^2 + 1$ and $1$ have the same number of zeros inside the circle $|z| = \frac{2}{3}$ This means that all of the roots of $z^8+z^2 + 1$ lie in the annulus $\frac{2}{3} < z < \frac{3}{2}$

# A bit more on complex exponents

I could go on for multiple lessons about this topic, so I’ll limit myself to a few remarks about the values of $z^w$ where both $z$ and $w$ are complex numbers. A good way to study complex arithmetic is to justify each step by using the definitions presented in class.

Let $z = re^{it}$ be written in complex polar coordinates with $r > 0$ a real number and $t \in arg(z)$. Let $w = a + bi$

Now $z^w = (re^{i(t + 2k \pi)})^{a+bi} =$

Now switch to $exp(z) = e^z$ notation:

$= exp(aln(r) +(ln(r) b)i))exp(i(t+2k\pi)a -(t+2k\pi)b)$ where $k \in \{..-2, -1, 0, 1, 2,...\}$

Now this becomes:

$exp((a ln(r) -b(2k\pi +t)) + i(ln(r)b + a (t + 2k\pi)))$

1. if $b = 0$, the exponent is real and we are back to what we originally had. This becomes very interesting if $a$ is irrational; we then get an infinite number of values.

2. Note that we can get an infinite number of values for the modulus AND argument if both $b$ and $a$ are irrational.

Feel free to explore what you get for various values of $a$ and $b$ for a fixed $r$ and $t$.

Let’s do an example: $1^{\sqrt{2} + \sqrt{3}i }$ Here $r= 1$, $t = 0$, $a = \sqrt{2}, b = \sqrt{3}$ and $ln(r) = 0$. So we end up with:

$exp(-\sqrt{3}2k\pi) exp(ia 2k\pi)$ as $k \ \{..-2, -1, 0, 1, 2, 3...\}$ and so we end up with an infinite number of values such as:

$1, exp(-(\sqrt{3})2 \pi)exp(2 \pi \sqrt{2}), exp((\sqrt{3}) 2\pi) exp(-2 \pi \sqrt{2}), ..$

The set forms a sequence spiraling out from the origin and going infinitely far away. Note that the arguments are irrational multiples of $\pi$

# A Theorem about line segments

This is regarding problem 16 of section 1.3

There is a theorem called the Heine-Borel Theorem from mathematics that implies that, given a closed line segment and a collection of open disks that contain that said line segment, only a finite number of those open disks are required to cover that said line segment. This theorem is usually proved in real analysis classes and real analysis is not a prerequisite for this class.

We will need a tool for our proof and that is the least upper bound axiom for real numbers: if a non-empty set of real numbers has an upper bound, then it has a least upper bound. This is an axiom of mathematics and not subject to proof.

Now let $S$ be any line segment of finite length in the plane; we can assume that $S = \vec{a} + \vec{b}t, t \in [0,1]$ (think of what you did in calculus 3).

Now let $\cup_{j=1} D_j$ be any covering of $S$ by open disks. Now let $c$ be the least upper bound of all numbers between $0, 1$ where the subsegment $\vec{a} + \vec{b}t, t \in [0,c]$ can be covered by a finite number of the $D_j$ Since at least one of the $D_j$, say $D_1$, contains the point $\vec{a}$ such a $c$ exists. But then the point $\vec{a} + c \vec{b}$ lines in another $D_k$ so the segment $\vec{a} + \vec{b}t, t \in [0, c + \epsilon]$ lies in the union of a finite number of open disks (the number of the previous collection, plus one more). So $c$ must not heave been an upper bound at all…unless $c = 1$ and the segment terminated there.

So the entire segment can be covered by a finite collection of these open disks.

# About that “point at infinity”

The text says that a set $X$ has the point of infinity in its interior if $X$ contains some set $|z| > M$ (for some $M > 0$) in its interior. How does one think of this?

Imagine a continuous map $g:D \rightarrow C$ which is both one-to-one and onto where $D$ is the open unit disk ($\{z | |z| < 1 \}$)

where $g(re^{it}) = (\frac{t}{1-t})e^{it}$. That is, $g$ stretches each ray segment $[0,1)$ through the origin out “to infinity” while keeping the argument the same.

Now think of the “point of infinity” as being where ENTIRE UNIT CIRCLE $|z| = 1$ would go to…so in the preimage (domain) picture, the point at infinity is identified with the entire unit circle. The open sets containing it would be open rings bordering the circle “on the inside of the disk”.

# Studying the function f(z) = z^2

In calculus 1, studying the graph of $f(x) = x^2$ was relatively easy. The domain of the function is the real line and the range is the non-negative reals. So the graph lies in $R^1 \times R^1$ and therefore can be viewed on the plane.

In calculus 3, we sometimes showed graphs in 3-space; for example the upper hemisphere of $x^2 + y^2 + z^2 = 1$ serves as the graph of $f(x) = \sqrt{1-x^2-y^2}$. the domain is a subset of the plane and the range is a subspace of the real line, hence the graph “lives” in 3-dimensions.

But the situation is very different in complex variables; the domain of the function is a subset of the complex plane (possibly all of the complex plane) and the range is also a subset (possibly all) of the complex plane. So the graph “lives” in a space that has 4 “real” dimensions (“real” as in “real numbers”…4 dimensions as in the vector space $R^4$. Yes, I am being pedantic because there is such a thing as “complex dimension”.)

So, how might we view a complex valued function with a complex domain?

One way might be to borrow what we learned about vector fields in calculus 3.

For example: $f(z) = z^2$ might be viewed as $f(x+iy) = (x+iy)^2 = (x^2-y^2) + i2xy$ which can be graphed as the vector field $F(x,y) = (x^2-y^2, 2xy)$

Here is such an attempt:

You can see the strangeness in the left half plane (where $Im(z) < 0$) and there is a reason for that.

To see that is going on, write $z = re^{it}$ where $t = arg(z)$ and $r = |z|$ then $z^2 = r^2e^{i2t}$. So the argument gets doubled and the modulus gets squared. Hence the vectors "rotate" as you go around the the x-axis and the lengths get very short near zero and very long when they are further away from the origin.

Exercise: let $u(x,y) = x^2-y^2, v(x,y) = 2xy$. Now calculate the following partial derivatives: $u_x, v_y$ and compare. Calculate $-u_y, v_x$ and compare. Now calculate $u_{xx} + u_{yy}$. What do you notice?

Another way: look at the image of certain subsets of the complex plane. Write $z = re^{it}$ and note that $(re^{it})^2 = r^2e^{i2t}$. Now note what happens: if we have the circle $|z| = a$ where $a < 1$ then those points are taken to a circle of radius $a^2 < a$ and points of a circle $|z| = b > 1$ are taken to a circle of radius $b^2 > b$. So the map squeezes circles inside of the unit circle and expands circles outside of the unit circle (circles understood to be centered at 0 ).

Now look at rays emanating from the origin; these rays have constant argument. An argument $t$ gets sent to $2t$ so: under $f(z) = z^2$ a ray of, say, argument $\frac{\pi}{6}$ gets sent to a ray of $\frac{\pi}{3}$, $\frac{\pi}{3}$ gets sent to $\frac{2\pi}{3}$ and $\pi$ gets sent all the say to $2 \pi$. So you can see that the rays get rotated by different amounts as the argument is doubled.

You can also see that half of the complex plane gets “stretched” to the entire complex plane; here is a relevant pictures (I am showing the half plane $Re(z) \geq =0$ and now it gets stretched to the entire complex plane; half-circles and rays in the first picture get taken to the circles and rays of the same color in the second picture.

I’d like to stress that the “rays of constant arguments” get rotated by different amounts; this is NOT a “rigid rotation”.

A “rigid rotation.” Consider the function $f(z) = (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}i)z$. Now in polar form: $f(re^{it}) = e^{\frac{\pi}{4}} r e^{it} = re^{i(t + \frac{\pi}{4})}$ which IS just a rigid rotation of $\frac{\pi}{4}$. This map does not change the absolute value but does add $\frac{\pi}{4}$ to the argument.

Exercise: show that $f(z) = (w_0)z$ (where $w_0$ is a complex constant ) has the following effect: $|f(z)| = |w_0||z|$ and $arg(f(z)) = arg(z) + arg(w_0)$. This is just a “rotation and a stretch” or “rotation and a dilation”.