Hello I recently completed the final exam for the "Daily Cause of Complex Analysis" course, and I successfully passed. However, when I downloaded my Level 1 certificate, I noticed that it appears to have some kind of defect.

I have attached both the certificate and a screenshot of my progress window for you reference . Could you kindly investigate this issue and provide assistance in resolving it?

Thank you for your support.

Attachments:

Captura de tela ...png

b7ca7e05-2a39-46...pdf

Question About Deforming Contours

Thanks for your response. I will need to read this over carefully.

I completed the exam for "Introduction to Complex Analysis" awhile ago. The download link for the level 1 certification in the online course is active, but it doesn't work. Can this be fixed, or is there an alternative way to get the certificate?

For Lesson 13 Problem 4, I can't see how the path for the horizontal section is rho(t)=1-2t with 1>t>-1 as stated in the solution. The path has a nonzero imaginary component. Isn't the path best described by rho(1)=1+it with t going from1 to -1?

Thanks so much for your reply

I've written a large Notebook. The output and input cells are interspersed. I'd like to have all the output together in 3 Sections with Section names, say "General Results", "Case A Findings", and "Case B Findings". I'd like the 3 Sections to have Subsections, and the Subsections to have Subsubsections. All the output cells can be either after the end of all the input cells or in a separate Notebook. How can that be coded?

I know how to generate a new Notebook with Sections, etc. For example:

nb = CreateDocument[{ipSection = TextCell["Input Data", "Section"], 

   aSection = TextCell["General Results", "Section"], 

  bSection = TextCell["Case A Findings", "Section"]},

  WindowSelected -> True, 

  WindowTitle -> 
   "Analysis Created on " <> DateString[Today],

  NotebookFileName -> "Analysis_" <> ToString[Today]]

However, I don't know how to "Print" output into the new notebook (nb in the example). For example, if I have an output plot generated using ComplexPlot3D[...], how do I get that into the Notebook nb in a specified section (like bSection)? I found ResourceFunction["PrintAsCellObject"] but I've been unable to use it to do what I just described. I've also looked at Write, etc. and have not been able to develop any code that works. The simplest thing would be to direct all the output to "Print" at the end of all the input in designated Sections and Subsections. Are there some settings in Format > Option Inspector ... that might help?

I agree with your answers to questions 5 and 6. I think you intended to show that the questions were misgraded regardless of how you answered them, but I think you ended up just repeating your answers.

Regardless, I agree with your answers as shown.

The grader must have been changed and broken.

Here are my reasons for agreeing with you. Question 6 presented before Question 5.

So the answer to Question 6 should be "Yes," as you chose, and the answer to Question 5 should be "No,", also as you chose.

In the solution to Chapter 9 Exercise 5 it is stated "The function is a conformal transformation at all points where it is analytic and where f'(z)!=0. " What does the notation f'(z)! mean?

I don't see how the quotient rule is being applied here to get the derivative of (z^2-1)/((3+z)(z-2i)).

Can anyone suggest what I am missing? .

Thanks! Makes sense in context.

The code in the D[] operator omits the (z+3) factor in the denominator of the function in the question. The solution is correct, if they meant the function (z^2-1)/(z-2i).

You might enjoy "The origin of the refractive index," Chapter 31 of Volume 1 of the Feynman Lectures on Physics. You can find it online. It seems Derek Muller is following Feynman's treatment.

Note also that Rayleigh scattering is treated in the following chapter.

Feynman's lectures are amazing. I find this part of his work especially so.

Hi Mitchell,

Here are some connections. Connections to other fields may require some background for a full understanding, such as a course in the field. First a definition:

Def: $f(z)$ has a pole of integer order $k>0$ at $c$ if $(z-c)^kf(z)$ is holomorphic at $c$ but is not holomorphic for smaller powers $k$. Example: The following has poles of orders $1,2,3$ at $z=10,20,30$ respectively: $$f(z)={(z-40)^4 \over (z-10)(z-20)^2(z-30)^3} \,.$$

A number $c$ is a pole of order $k$ of $f(z)$ if it is a zero of multiplicity $k$ of $1/f(z)$. If we express the example above as a product, zeros and poles are distinguished by the sign of the exponent; otherwise, they are algebraically similar: $$f(z)={(z-40)^4 (z-10)^{-1}(z-20)^{-2}(z-30)^{-3}} \,.$$

Generalized mathematical significance (hope this isn't too abstract): Suppose we have a continuous object, such as a function $f_a(z)$ that varies continuously with one parameter/coefficient (or more) represented by $a$. And suppose there is a discrete quantity that is calculated from $f_a(z)$, such as the number of zeros or number of poles. Finally suppose that the discrete quantity can be proved to vary continuously with $a$. Then the discrete quantity is an invariant of $f_a(z)$. That is, it's constant.

Contrariwise, if the discrete quantity changes (is discontinuous), something interesting is probably happening with the functions $f_a(z)$ at the value for $a$ at which there is a discontinuity.

For instance, a meromorphic function on the Riemann sphere has a finite number of zeros and the same number of zeros ( $Z$) as poles ( $P$) counted with multiplicity and including poles/zeros at infinity. Thus $Z-P=0$. It is constant. You cannot change the number of zeros of $f_a(z)$ without also changing the number of poles in the same way (assuming $f_a(z)$ is meromorphic for all $a$). Example: $f_a(z)=z^2+a$ has two zeros because it has a pole of order $2$ at infinity (and vice versa).

More examples:

I suppose it's clear that solving equations has many applications, and the number of solutions has implications for the problem in which the equation arises. A solution to $f(z)=c$ is a zero of $g(z)=f(c)-c$.

I suppose it's also clear from the complex analysis course that the poles themselves are important in integration. Perhaps the number of them is not so significant, but it might sometimes be helpful to know how many there are.

An application: I suppose periodicity has been an important topic in mathematical physics and therefore in mathematics since humans noticed things like days, months, and years recur periodically. A periodic function has infinitely solutions to $f(z)=c$ if it has one. If $f(z)$ has infinitely many zeros, then it must have an essential singularity on the Riemann sphere (examples: $e^z$, $\sin z$, $\sin(1/z)$). Periodic functions must have this feature.

In control theory, the zeros and poles of the transfer function determine the behavior of a linear system. See MIT notes (read $2{du\over dt} + u$ for the RHS of the example at the top of page 2) or CalTech notes (read $e^{st}$ in the RHS of the equation between (6.6) and (6.7) on page 142).

I ask for an explanation.

I successfully plotted 5 points on the complex function f[z]=Sqrt[(z+1)(z-1)], with Show[ComplexPlot3D, Graphics3D].

If I use the same procedure and plot the point p1=(1, 1, Abs[g[z1]] ), with z1=1+I, on the function g[z]=Log[z] the program does not work. Why? I attach my notebook.

Attachments:

points onto Comp...nb

A thousand thanks. I wouldn't have solved it without your help.

You integrate on the boundary (Circle[]) and sum the residues in the interior (Disk[]):

ResidueSum[{z(z - 1)/((z - 2) (z - 3)), {Re[z], Im[z]} \[Element] 
   Disk[{0, 0}, 5/2]}, z]

By the Residue Theorem, you need to multiply the residue sum by $2 \pi i$ to get the value of the integral.

Hi Mitchell,

A Circle[] represents the boundary of a Disk[]. If {x, y} is a in point "in" the set represented by the Cirlce[], then the point is on the circumference and not in the interior of the circle. A Disk[] represents the region consisting of the circle and its interior. Since we want the sum of the residues over the interior, we need to sum over the Disk[].

This distinction between a circle and disk was not made when I took geometry in school. In Euclid a circle is "a region contained by a [curved] line called the periphery." That is, it was the same as a disk. It was a little strange when the distinction was made, I think around the time I got to line integrals and Green's theorem. When we got to surface integrals a bit later, I ran into a similar distinction between sphere (hollow surface) and ball (solid, equal to the sphere plus its interior). (They are Sphere[] and Ball[] in WL.)

I think I tried Element[z, Disk[{0, 0}, 5/2]], but it didn't work. I guessed I needed to specify the coordinates since Disk[] is two-dimensional in the two-dimensional real plane. One could specify the complex disk with Abs[z] < 5/2, which is how the examples in the docs are set up.

This calculus class included complex analysis introduced us to numerous formulas and theories. Are there any comprehensive collections of all the formulas and theorems related to calculus? It can be challenging to remember everything, so having a resource that compiles them all would be incredibly helpful.

The section "Bonus materials" of the course is a super-shortened condensate of the course with all the main formulas.

You are right, thank you. I'll have it fixed...

Question on notation in Section 12. Should the equation read the contour integral over C1 equals the contour integral over C2 (it says it equals the contour integral over negative C2)? I believe that C2 is defined as being in the clockwise direction, so the negative sign will naturally arise when the integral is done in the counterclockwise direction.

There is a typo in Problem 6 on Quiz 4. The first answer should read: "The integral does not exist." (and not "The integral does not").

Are branch cuts in a unique location? In Lesson 5 exercise 3 we are asked to "describe the branch cuts" of the function Sqrt[(e^2+1)]. Given the definition from Wolfram Mathworld "A branch cut is a curve (with ends possibly open, closed, or half-open) in the complex plane across which an analytic multivalued function is discontinuous. For convenience, branch cuts are often taken as lines or line segments." The solution states:

I can see from the figure that there are discontinuities for any line parallel to the Im axis for x>0. Is the branch cut where these discontinuities start or can the branch cut be anywhere x>0?

In Section 11 of the Lesson, shouldn't the first definition of a complex integral be the integral of f(z)dz over the contour (and not f(x)dx over the contour).

In Example 5.3, shouldn't the text read: "In other words, the branch cut corresponds to either z purely imaginary or to z real between -1 and 1."

Shouldn't the Caption to Fig 9 in the book be: "Fig. 9 Vector Plot of z^2" and not "Fig. 9 Vector Plot of z^2 + 1 = z^2"

If f(z)=2+2z+ Sum(z^n), where n = 2,3, ..., Infinity, the output of SeriesCoefficient[f[z],{z,0,n}] is:

1 n>1

2 n==0||n==1

0 True

What does "0 True" mean?

I've worked the quiz problems and they all make sense to me, except for one.The answer I get for Quiz 8, Problem 5 is present among the offered choices, but it's marked wrong by the grader.

I just tried the other three choices and found the one the grader likes, but I don't understand why.

Anyone else experience this -- or know why the answer marked correct is correct? I can get the answer it marks correct if I integrate over only half the unit circle.

As I said, the answers to all the other quiz problems make sense to me.

Hello,

I have some questions about quizzes 8 and 9, specifically problems 5 and 2. For problem 5, I selected option C, and for problem 2, I chose option B. I cross-checked my answers using mathematical references and other reliable sources, which confirm that my answers are correct. However, the official answer key lists different alternatives. Could you please clarify if there might be any errors in the answer key?

Thank you for your assistance.

Best regards, Ricardo

Hi;

In working with the properties of Harmonic Functions, how is the mean value calculated when you are given the formula of (x,y) along with the radius and center of a circle? I have gone through the course materials up to include unit 18 and have not found anything to help me set-up the problem. Thanks, Mitch Sandlin

Thanks to another student, I found this article in the Mathematica Journal: "Domain Coloring on the Riemann Sphere. I would like to use the code. But simply downloading the article as a notebook fails to give me anything other than the ability to look at the pictures.

I think I need to download and instal a package called complexVisualize.m.

Are you able to tell me where to find that file?

I am using this course to both extend my Mathematica knowledge and refresh my Complex Analysis.

Thank you. I'll have it fixed.

Hi;

Is there a straight-Forward method of producing a Stereographic Projection of a complex number? The examples given in our discussion materials seems extremely convoluted with several complex plots joined together with possibly some primitives added in.

Thanks,

Mitch Sandlin

Perhaps the contributed function RiemannSphereComplexPlot[] -- a 3D rotatable Riemann sphere version of ComplexPlot -- is what you are seeking: https://resources.wolframcloud.com/FunctionRepository/resources/RiemannSphereComplexPlot/

Hi,

I'm not exactly sure what you want, but here are formulas for the various elements of the projection. You can put them together in graphics as you see fit:

(* elements *)
pole = {0, 0, 2};
numberPt = {Re[z], Im[z], 0};
projectionPt = { (* main formula: proj. onto R. sphere *)
  (4 Re[z])/(4 + Abs[z]^2), 
  (4 Im[z])/(4 + Abs[z]^2), 
  (2 Abs[z]^2)/(4 + Abs[z]^2)};
riemann = Sphere[{0, 0, 1}, 1];
plane = InfinitePlane[{{0, 0, 0}, {1, 0, 0}, {0, 1, 0}}];

 z = 2 - I; (* need to give z a value *)
 Graphics3D[{
   Opacity[0.6],(* change style as desired *)
   Point[{pole, numberPt, projectionPt}],
   Line[{pole, numberPt}],
   riemann,
   plane},
  PlotRange -> {{-3, 3}, {-3, 3}, {0, 2}},
  FaceGrids -> { (* may omit *)
    {{0, 0, -1}, {Range[-3, 3], Range[-3, 3]}}}
  ]

Changing FaceGrids to the following will highlight the real and imaginary axes:

FaceGrids -> { (* may omit *)
  {{0, 0, -1},
   {Range[-3, 3], Range[-3, 3]} /.
     0 -> {0, AbsoluteThickness[1]}}}

Omitting FaceGrids gives a plain plane below the sphere. Omitting plane and using FaceGrids gives a nice look, too.

Hi Michael,

In my Complex Analysis textbook (Bak & Newman), the Riemann Sphere has unit diameter and sits above the complex plane tangent at $z=0$ (the origin), just as you say. It has the nice property that the equator maps to the unit circle.

In my code, I used a sphere with a unit radius, since this seemed to be quite common in a google search. I thought maybe things had changed since I took the course. The equator in this model is nothing special, though.

I have seen before the Riemann Sphere presented as the standard unit sphere (unit radius centered at the origin) so that the complex plane cuts it in two. This model seems less common to me. It does have the property that the equator is the unit circle.

All three models are effectively equivalent: Any proof or explanation based on one can be translated to another, changing projection formulas if necessary. In all three, the south pole corresponds to $z=0$ and the north pole to $z=\infty$, which is convenient.

Because of this, it seems more a matter of convention than correctness to me. I personally prefer the tangent-sphere models to the sphere centered at the origin. (And I like the unit diameter one better than the unit radius one, but I went with the peer-pressure of google in my previous reply.) Possibly the one centered at the origin is easier to deal with pedagogically.

Hi Tingting,

The only problem is if the "north pole" $N$, the point that you connect with $z$, lies in the complex plane. If it's not in the complex plane, then the line through $N$ and $z$ intersects the sphere in another point $P(z)$, which we may define as the projection of $z$ onto the sphere.

If the sphere is tangent to the complex plane at $N$, then all lines through $N$ and $z$ are tangent to the sphere at $N$. The projection is undefined. (Or all numbers $z$ are projected onto $N$, which is pretty useless.)

Right! the projected points end up on the other side of the sphere. And you were right before: If you use rays from the north pole $N$, then you get what said in your first reply, and the north pole has to be above the $z$ plane to get a second intersection point. (And if one uses the line segment from $N$ to $z$, then the whole sphere has to be above, or tangent to, the $z$ plane. I was used to lines, which avoids concerns about betweenness.)

Thank you, This is nice and will, I hope, give me some basis for experimentation. I would like to define my own shapes on the sphere and see the mapping not the plane.

You are correct, I'll have it fixed. Thank you!

Lesson 9 states "for the limit of a complex function to exist, it must exist no matter the direction by which it is approached." Is a necessary and sufficient condition for this that the limit exists and is the same for all approaches along straight lines? Is it possible for the limit to exist and be the same for all approaches along straight lines but for that not to be the case when approaching along some curve that is not a straight line?

I think @Marco is right. If we are considering arbitrary functions of a complex $z$, here is an example based on a real two-variable example:

$$f(z)=\frac{\left(z-z^*\right) \left(z^*+z\right)^2}{\left(z-z^*\right)^2-\left(z^*+z\right)^4}$$

Along lines through $z=0$,

$$f\bigl((a+ib)\,t\bigr)=-\frac{2 i a^2 b}{4 a^4 t^2+b^2} \,t$$

approaches $0$. Along certain parabolas through $z=0$,

$$f(t a + i b t^2)=-\frac{2 i a^2 b}{4 a^4+b^2}$$

is constant and so has a nonzero limit if $a$ and $b$ are nonzero.

Here is a visual:

ComplexPlot3D[
 ((z + Conjugate[z])^2 (z - Conjugate[z])) /
  ((z - Conjugate[z])^2 - (z + Conjugate[z])^4)
 , {z, -1 - I, 1 + I}, PlotRange -> {0, 1}, 
 MaxRecursion -> 6]

Michael, I've written some code that enhances the ComplexPlot3D display of the function

f[z_] := ((z + Conjugate[z])^2 (z - 
      Conjugate[z]))/((z - Conjugate[z])^2 - (z + Conjugate[z])^4)

The code is in the attached Notebook, file ComplexLimitCounterExample_16nov24.nb .

From the plots, it appears that the values of function f have only two arguments (angles with the real axis) and that they are determined by whether Im[z} is positive or negative. I have not analyzed f to confirm this. Am I correct? Do you see a simple analysis to determine this?

BTW, "Calculus Volume II" by Tom M. Apostol, Copyright 1962 by Blaisdell Publishing Company has on page 70 a simple real-valued function of two real variables that is not continuous at 0 even though approaching 0 along either axis has limit 0. However, when approaching 0 along the line x = y, its limit is 1/2. Here is the function:
f(x,y) = x*y / (x^2 + y^2) if (x,y) != (0,0) and f(0,0) = 0.

Attachments:

ComplexLimitCoun...nb

No, (z - z*) = 2 i Im(z)

Hi Gerald,

For some reason, I wasn't notified of your replies. Since Study Group threads get long, I didn't happen to see them until today. Nice notebooks, btw. You asked:

From the plots, it appears that the values of function f have only two arguments (angles with the real axis) and that they are determined by whether Im[z} is positive or negative. I have not analyzed f to confirm this. Am I correct? Do you see a simple analysis to determine this?

I think the following shows it:

ClearAll[f]
f[z_] := ((z + Conjugate[z])^2 (z - 
      Conjugate[z]))/((z - Conjugate[z])^2 - (z + Conjugate[z])^4)

ComplexExpand[f[z], z, TargetFunctions -> {Re, Im}] // Simplify

(*  -(2 I Im[z] Re[z]^2)/(Im[z]^2 + 4 Re[z]^4)  *)

Note the squares on Re[z] and Im[z] except for the factor of Im[z] in the numerator. The argument of f[z] is the same as the argument of -I Im[z], which is -Pi/2 when Im[z] > 0 and Pi/2 when Im[z] < 0.

In a different reply, you remarked:

On a side note, if a real valued function of a complex variable has a derivative at any point, that derivative must be zero.

Right. It was this sort of thing I had in mind when I prefaced my example with, "If we are considering arbitrary functions of a complex variable $z$..." [emphasis added]. The function f[z] is not complex-differentiable.

What can you say about physical analogues to Residue Theory -- how it applies to physics, electromagnetism, etc.? I found myself completely disconnected to what a "residue" means -- how a number of singularities has physical significance. What is the significance of the orthogonality of harmonics? As a professor, how to you create the engagement/pertinence of these abstract concepts, @Michael? I can search and ask the AIs, but I wanted to know what you and @Marco can say personally. Thank you.

I haven't taught Complex Analysis yet. So I haven't had to think about how to present the subject to students. You might be more disappointed in this opinion: Not every mathematical concept needs to anchored in, or even connected to, a physical phenomenon or application. When I was a physics major in college, my interest in complex analysis and in math in general was in understanding how functions work and to have better skills for solving the math in scientific problems. It was clear from my physics professors that such skills would be of value in physics.

Residues let you calculate integrals; indeed, even integrals from real analysis, which is a surprise. Calculating integrals was a skill valued by me and my fellow physics majors. You might think Mathematica would take care of all that. But it doesn't. Sometimes the integrand contains a general function that has no explicit formula but about which there are only some assumptions. Residues, branch cuts, integration by parts, and other tools may help say something about the integral. Other times, the understanding I have obtained from calculus, complex analysis, and differential equations has allowed me to help people (or myself) when Integrate[] has trouble. Sometimes I can figure out workarounds. Understanding complex integration sometimes has helped solve issues with NIntegrate[].

So: How do I create engagement? That invites a long answer. I'll boil it down to a quote (from faulty memory). Eva Brann, basing a question on the root meaning of "to teach," which is "to show," asked what is the best thing a teacher can show?: "How they think."

What residues mean to me: - The integral of $(z-a)^k$ around a simple closed path is zero unless $k=-1$ and the path goes around $a$, in which case the integral is $2\pi i$. - This fact is due to Green's theorem, the Cauchy-Riemann equations, and a simple calculation for the path $z=a+e^{it}$. - This means that if $f$ is analytic or has a Laurent expansion $f(z) = \sum\limits_{k=n}^\infty {c_k (z-a)^k}$, and we integrate around a path that contains $a$ and stays within the radius of convergence, then the integral of each term of the series vanishes except for the $k=-1$ term; and the integral is determined by the residue (times $2\pi i$). This is the key for me. - If the path goes around several poles, the integral is determined by the sum of the residues. - Dimensional analysis: Given that the integral $\int (z-a)^k\;dz$ is path-independent, the value is constant for paths that go around $a$. The integral also scales by like the $k+1$ power: If the path is a circle centered $a$, then the integral around a circle of radius $2r$ is equal to $2^{k+1}$ times the integral around a circle of radius $r$. Since the integrals are equal, the integral must be zero unless $k=-1$. This is one way to look at why the $-1$ power is the only significant one. It also suggests why expanding a function in a power series is revealing. - The coefficient $c_k$ of the series expansion of $f(z)$ at $a$ may be computed from ${1 \over 2\pi i} \int f(z)(z-a)^{-k-1}\,dz$. (This may be done efficiently in NIntegrate[] with the "Trapezoidal" strategy.) Multiplying by $k!$ yields the value of the derivative $f^{(k)}(a)$. In other words, dividing the Laurent series of $f(z)$ by $(z-a)^{k+1}$ shifts the coefficients of the series so that $c_k$ becomes the residue. - The radius of convergence is equal to the distance from the center to the nearest pole (other than $a$). So the path staying within the radius of convergence means it does not go around another pole and pick up another residue.

I'm not sure of the context for "orthogonality of harmonics." In AC power, the average power is the inner product of the voltage and current, which can be broken down into a linear combination of harmonics. If they are composed of different harmonics, then the average power is zero. If they have nonzero components for the same harmonic, then the average power will be nonzero. (If you're talking about Fourier series. In general, an orthogonal basis is golden, in many different topics in mathematics; just as square is in carpentry and woodworking.)

All code can already be seen, including that for images and animations. You just have to double click on the bracket that hides the input, identifiable by the little upward-pointing arrow:

Double clicking will expand the code:

In Lesson 3, why does the Postfix instruction //Labeled[#,Text[ ...]& result in just one Text for the entire row rather than a separate text for each of the 3 ComplexPlot3D[...] entries in the Row[...] for the following code?

Row[{
ComplexPlot3D[z,{z,3},PlotRange->{{-3,3},{-3,3},{0,6}},BoxRatios->{1,1,1},PlotStyle->Directive[Opacity[0.8`]],AxesLabel->{Style[ToString[Re[z],TraditionalForm],12],Style[ToString[Im[z],TraditionalForm],12],Style[ToString[Abs[z],TraditionalForm],12]},ImageSize->160],
ComplexPlot3D[Conjugate[z],{z,3},PlotRange->{{-3,3},{-3,3},{0,6}},BoxRatios->{1,1,1},PlotStyle->Directive[Opacity[0.8`]],AxesLabel->{Style[ToString[Re[z],TraditionalForm],12],Style[ToString[Im[z],TraditionalForm],12],Style[ToString[Abs[Overscript[z, _]],TraditionalForm],12]},ImageSize->160],
ComplexPlot3D[1/z,{z,3},PlotRange->{{-3,3},{-3,3},{0,6}},BoxRatios->{1,1,1},PlotStyle->Directive[Opacity[0.8`]],AxesLabel->{Style[ToString[Re[z],TraditionalForm],12],Style[ToString[Im[z],TraditionalForm],12],Style[ToString[HoldForm[Abs[1/z]],TraditionalForm],12]},ImageSize->160,PlotLegends->Automatic]
}]//Labeled[#,Text[StringJoin["Fig. 4. Plots of the functions ",ToString[z,TraditionalForm],", ",ToString[Overscript[z, _],TraditionalForm],", ",ToString[1/z,TraditionalForm],"."]]]&

I think I see the answer to my question. The Postfix code is just a function; no Map (/@) is specified.

For figure 2 lesson 3, is the caption correct?

Trying to catch up here. Surely this is a simple question. When plotting Re[z] and Im[z], shouldn't these plots be confined to either the {z, Re[z]} or the {z,Im[z]} planes? I don't see why these plots appear to be surfaces in 3 dimensions.

Thanks!

Attachments:

14381729.nb

I am wondering about the history of complex numbers, Girolamo and Rafel, in 16C. How did they find the need, and what profession both were in? Did mathematician careers exist in 16C?

Thxs, Marco.

I am wondering how you got into complex number analysis. What trigger made you enter this field?

In the European tradition of philosophy, Plato famously asked (demanded?) a knowledge of mathematics from his students. At that time, I think, mathematics was considered as part of Logic which was an essential component of philosophy.

The Pythagorean School (much earlier than Plato) also took mathematics very seriously, but the school looks quasi religious by today's (western) standards.

Can you comment on the significance or the application of the Riemann sphere concept? Why should it matter how lines or circles in the complex plane map to the sphere?

Wow. That needs to find itself into some SciFi story about space travel. Love it. Thanks.

Hi Joseph,

The Riemann sphere is a way to understand the role of infinity in complex analysis. To understand functions of a complex variable is a mathematical application, but it is important and the one for which the Riemann sphere was invented. To fully understand it, one needs to know some (differential) topology, which I will not attempt to explain. I will state some facts and hope they will be intelligible enough that they will be motivating. The upshot is this: The Riemann sphere is a really good, accurate model for understanding analytic, complex-valued functions. Every point on a sphere looks the same by symmetry. Via the Riemann sphere we can define limits, derivatives etc. at infinity and treat infinity just like any other complex number.

Every nonconstant rational function $R(z)=P(z)/Q(z)$, with $P$ and $Q$ polynomials, maps every point on the Riemann sphere to another point on the sphere; and further, every point is the value of $R(z)$ for at least one point $z$.

• $R(z)=1/z$ maps $0$ to $R(0)=\infty$ and $\infty$ to $R(\infty)=0$.
• ${1 + 2z^2 \over 12 - 3 z^2}$ maps both $\pm2$ to $R(\pm2)=\infty$ and $\infty$ to $R(\infty)=-2/3$.

To get to another part of your question, it turns out that the only differentiable symmetries (a.k.a. "diffeomorphisms" in differential topology) of the Riemann sphere as a model of complex analysis are fractional linear transformations, a.k.a Möbius transformations: $$z \mapsto {az+b \over cz + d}$$ Fractional linear transformations map circles and lines in the complex plane to circles and lines. Circles and lines in the complex plane correspond to circles on the Riemann sphere, and so the Riemann sphere gives a unified look to them. It's not incorrect to think of a line as a circle with an infinite radius in this context.

Symmetries play an important role in understanding how things are related. For instance, $1/(z-2)$ is just the function $1/z$ translated over two units. What one knows about $1/z$ corresponds to the same facts about $1/(z-2)$ translated over. Likewise, knowledge about $1/z$ corresponds to the same facts about $(1+i)/z$, but with function values scaled by $1+i$. Fractional linear transformations, then, are the ways that one can translate facts about the derivatives etc. of a given function $f(z)$ to other functions. If $f(z)$ and $g(z)$ are related by such symmetries, then they are essentially the same function, with locations and values connected by the symmetries.

A particular class of functions that is important in this context is the class of functions that can be represented by power series. You should get to that later in the course.

You will also see other ways that infinity is important in this course (poles, residues, to name two). Going beyond the course, in topology I learned why the first type of integral below is easy and taught in first-year calculus and why the second has books written about them: $$(1)\ \int {dx \over \sqrt{\text{quadratic polynomial}}} \qquad (2)\ \int{dx \over \sqrt{\text{cubic polynomial}}}$$

The characterization of fractional linear transformations in terms of lines and circles makes even more geometric (and visual) the projection connecting the Riemann sphere and the complex-number plane. That possibility of visualization also makes WL/Mathematica a particularly good tool to use to explore complex analysis.

Hi Gerald,

I learned these things back in the 1980s, some facts as an undergraduate but mainly in graduate school from lectures. The professors sometimes recommended books as ancillary materials, but only a few used books as textbooks and followed them. Further, I've been at an institution teaching only 1st and 2nd year students for 30 years, so I haven't taught this or related material for a long time. Nonetheless, I can give you some recommendations, but I personally have not field-tested them, so to speak.

Integrals of the form (2) first showed up for me as an undergraduate in physics. Integrals with the square of 3rd and 4th degree polynomials are called elliptic integrals. The arc length of an ellipse can be expressed with such an integral, hence the name, and they first came to the attention of mathematicians in the mid 18th century. They were studied somewhat intensely in the early 19th century by Gauss, Jacobi, Legendre, and in particular by Abel, who extended the scope to include integrals involving square roots of higher degree polynomials. These are called hyperelliptic or "Abelian* integrals. Much knowledge of them was developed before the Riemann sphere or topology was even invented. These facts give you some keywords to google, but they also warn you that you will find a lot of information about (2) not relating to my remarks above.

I came to know more about elliptic integrals from a sideways direction. I was studying number theory and algebraic geometry in grad school; in particular, "elliptic curves" was a hot topic. They come from the square-roots in the elliptic integrals, that is, curves of the form $y^2 = \text{cubic or quartic}$. In the 1980s, they were thought to hold a key to proving Fermat's Last Theorem (which turned out to be the case). They also yielded a powerful factorization algorithm, which was important because of the then relatively new RSA encryption algorithm. It's truly amazing the connections the integral (2) has to seemingly unrelated branches of mathematics.

It turns out there is a book that "does it all", or almost:

Elliptic Curves: Function Theory, Geometry, Arithmetic by McKean and Moll

The book was recommended to me by a friend, and on that basis, I am recommending it here. It gives an explicit construction of the differential manifold structure on the Riemann sphere, which is fairly simple, considering. It also gives a definition of manifold. So it covers everything that is necessary. However, it is question in my mind whether, having given someone the needed tool, they will know why, when, and how to use it. But it's less likely to be a problem than the general case, because the goal is to understand this manifold, namely the Riemann sphere, not manifolds in general. That aside, it covers elliptic integrals, elliptic functions, elliptic curves (though not the connections to Fermat's Last Theorem and factorization), and a few other things.

For differential topology, it's a bit harder to suggest a book. Many assume the reader is familiar with topology, especially manifolds. General introductions to topology include manifolds and much else that isn't strictly necessary for the question at hand. The following was my undergraduate textbook. I remember liking it and using it as a reference in grad school from time to time. And it seems to have the right scope, at least through, say, Chapter 5, which covers differentiable manifolds:

Singer and Thorpe, Lecture Notes on Elementary Topology and Geometry.

This Math.SE Q&A contains some good recommendations:

https://math.stackexchange.com/questions/46482/introductory-texts-on-manifolds

These two have a good reputation:

• L. Tu, An Introduction to Manifolds (includes differentiable a.k.a. smooth manifolds†)
• John M. Lee, Introduction to Smooth Manifolds

†Note: Usually there is a slight difference between these two; however, every smooth manifold is a differentiable manifold.

I hope that helps.

You are right, I'll have to fix the exercise. Thank you

You are correct, thank you!

Thxs, Gerald for the updated references. Do any books cover aerospace applications that use complex numbers?

So, I asked this question during the lecture and Marco did not have an answer for it. I had never heard of the stereographic projection of complex numbers. I was wondering if anybody knows of a practical application for it. It seemed to me that it might be used in modulation/coding theory or maybe image recognition problems but I'm just shooting in the dark. Anybody know?

This method is used to produce a map from a sphere to a cartesian plane. Say for example to show a flat map of the spherical Earth.

Oooo, bro set x = s + t so that he can cancel the imaginary numbers in certain conditions. Noice!

Please look at the 3rd question in Quiz 1. The question asks whether a given point is stereographically projected into the lower or the upper hemisphere of the Riemann sphere. In fact, the magnitude of the point is exactly 1, so that the given point lies on the equator. But that's not an available answer.

So, what's the right response?

yeah, I noticed that too. I picked North since the equator was not available. It said it was the wrong answer.