Hello Joseph,

Thank you for reaching out. I will follow up with you outide this post.

Best,

Christine Owens Wolfram U Project Manager Wolfram U

You are right... Thank you!

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).

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).

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.

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?

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

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 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.)

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.