\(\renewcommand{\Re}{\mathop{\textrm{Re}}} \renewcommand{\Im}{\mathop{\textrm{Im}}} \)

Journal for Math 407, Spring 2012

Friday, May 4

The final exam was given, and solutions are available.

Thursday, April 26

This was the last class meeting for the semester (since next Tuesday is redefined as Friday). We had some general discussion about how the course fits into the broad panorama of mathematics, including the Riemann hypothesis.
Reminder: The final examination takes place early in the morning (7:30–9:30) of Friday, May 4. My office hours next week will be 2:00pm–3:00pm on Wednesday and Thursday (May 2 and 3).

Tuesday, April 24

We worked on Problems 8.29a, 8.30a, and 8.34b.
The assignment is to make a list of the important theorems and formulas that we covered this semester.

Sunday, April 22

I posted solutions to the quiz from last week.

Thursday, April 19

In class, we discussed analytic functions from the point of view of linear algebra; in particular, we discovered that the Cauchy–Riemann equations are equivalent to saying that the real-linear approximation of the function is given by a matrix that corresponds to a complex-linear transformation.
Here are three problems related to geometry to do for next time (not to hand in, for there is not time to get them back from the grader).

Problem 1.144, which says to show that \(\Re(z)\gt 0\) if and only if \(\left|z-1\right|\lt \left|z+1\right|\).
You should be able to find both an algebraic solution and a geometric solution.
Problem 2.133, which says to show that \[\frac{\sin(\theta)}{2} + \frac{\sin(2\theta)}{2^2} + \frac{\sin(3\theta)}{2^3} + \cdots = \frac{2\sin(\theta)}{5-4\cos(\theta)}. \] The “geometry” here is a hidden geometric series. Can you see how to use this formula to give a new solution to the first quiz problem from last time?
Problem 8.34, which asks for the equations of the curves in the \(w\) plane into which the straight line with equation \[x+y=1 \] is mapped by the transformations \(w=z^2\) and \(w=1/z\).
You should find that the two image curves are a parabola and a circle.

Part of the point of this assignment is to remind you to start reviewing for the final examination, which is scheduled for Friday, May 4, at 7:30 in the morning. The final exam from last semester is available, along with solutions.

Tuesday, April 17

In class, we worked on the following three problems; since we did not finish, these problems are a take-home quiz to hand in next time.

Show that \[ \int_0^{2\pi} \frac{(\sin(\theta))^2}{5-4\cos(\theta)}\,d\theta=\frac{\pi}{4}.\]
Evaluate \[ \int_{C_R} \frac{e^{iz}}{(z^2+1)^2}\,dz,\] where \(C_R\) is the triangle with vertices at \(-R\), \(R\), and \(iR\), and deduce that \[\int_0^\infty \frac{\cos(x)}{(x^2+1)^2}\,dx = \frac{\pi}{2e}.\]
Evaluate \[\int_{C_N} \frac{\pi }{z^2 \sin(\pi z)}\,dz,\] where \(N\) is a positive integer and \(C_N\) is the square with vertices at \((N+\frac{1}{2})(\pm1\pm i)\), and deduce that \[\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^2} = \frac{\pi^2}{12}.\]

Thursday, April 12

In class, we discussed the previous assignment, and we solved problem 7.49 by applying the residue theorem in a new way. We did not have time for a quiz, so plan for a quiz next time. Here are some problems to do for next time (not to hand in).

Problem 7.47, which asks for the value of \[\oint_C \frac{2+3\sin(\pi z)} {z(z-1)^2}\,dz \] when \(C\) is the square with vertices at \(\pm3\pm3i\).
The answer in the book has an error: the answer should be \(-6\pi^2 i\).
Problem 7.58, which asks for the value of \[\int_0^\infty \frac{1}{x^4+x^2+1}\,dx. \] Suggestion: By symmetry, the integral is one-half the integral over the whole real axis. Use a semi-circular path with diameter from \(-R\) to \(R\). You will need to compute residues at two simple poles.
Problem 7.76, which says to show that the sum of the residues of the function \[\frac{2z^5-4z^2+5} {3z^6-8z+10} \] at all the poles is equal to \(2/3\).
Suggestion: Factoring the denominator is hopeless, but what happens if you integrate the function around a circle of very large radius?

Wednesday, April 11

Notice that the answer in the book to Problem 7.78 is missing a factor of \(2\pi i\).

Tuesday, April 10

We discussed methods for finding residues at simple poles, and we worked an example of evaluating a real integral of a periodic function by converting the problem into a complex line integral over the unit circle.
Here is the assignment for next time (not to hand in, but be prepared for a quiz).

Problem 7.48, which asks for the value of \[\frac{1}{2\pi i} \oint_C \frac{e^{zt}} {z(z^2+1)}\,dz \] when \(t\) is a positive number and the path \(C\) is a square with vertices at \(\pm2\pm2i\).
Problem 7.52, which asks for the value of \[\int_0^{2\pi} \frac{\cos(3\theta)} {5+4\cos(\theta)} \,d\theta. \] This problem is similar to solved problem 7.13. Can you find an easier way to determine the residue at the third-order pole than the method shown in the book? The final answer to the problem is \(-\pi/12\).
Problem 7.78, which asks for the value of \[\oint_C z^3 e^{1/z}\,dz \] when \(C\) is the circle of radius \(4\) centered at the point \(1\).

Thursday, April 5

In class we worked on computing some integrals by using a tool that we subsequently learned to call the residue of a function at an isolated singularity.
The assignment for next time is to read Section 6.10 about the classification of singularities and to solve the following problems (to hand in next time).

Problem 6.96a,b, which says to expand the function in a Laurent series about the origin and to name the type of singularity for (a) the function \((1-\cos(z))/z\) and (b) the function \(e^z/z^3\).
Problem 7.40, which says to show that \[\oint_C \frac{\cosh(z)}{z^3}\,dz=\pi i \] when \(C\) is the square whose vertices are the points \(\pm2\pm2i\).
[The hyperbolic cosine function is defined in Section 2.6.]
Problem 7.44, which asks for the value of \[\oint_C e^{-1/z}\sin(1/z)\,dz \] when \(C\) is the unit circle centered at the origin.

Tuesday, April 3

The topic for class today was connections between series and integrals. We derived Taylor's formula from Cauchy's integral formula, and we started discussing Laurent series. Geometric series were a key tool.
The assignment to hand in next time is Problem 6.92, which asks for five different Laurent series for the rational function \[\frac{z}{(z-1)(2-z)} \] as follows: (a) the series valid when \(|z|\lt 1\) (which is just the ordinary Taylor series with center at \(0\)); (b) the series valid in the annulus where \(1\lt |z|\lt 2\); (c) the series valid in the region where \(|z|>2\) (the exterior of a circle); (d) the series valid in the region where \(|z-1|\gt 1\) (this series will be in powers of \((z-1)\) and \(1/(z-1)\)); and (e) the series valid in the punctured disk where \(0\lt |z-2|\lt 1\) (this series will be in powers of \((z-2)\) and \(1/(z-2)\)).
Remark: In part (d), observe that \(z-2\) can be rewritten as \((z-1)-1\), and in part (e), observe that \(z-1\) can be rewritten as \((z-2)+1\). The answer to part (c) in the book has a typographical error: the initial term \(-1/2\) should be \(-1/z\).

Thursday, March 29

The second exam was given, and solutions are available.

Tuesday, March 27

We reviewed for the exam to be given next time on Chapters 4, 5, and 6 (on those parts that we have covered). The assignment, of course, is to study for the exam.

Thursday, March 22

My upcoming office hour on Monday, March 26 will be 9:00–10:00 in the morning (instead of the usual afternoon time). This change is a one-time event due to a scheduling conflict.

In class today, we continued the discussion of convergence of series, and we covered the advanced version of the root test (using the limit superior). Here are some problems to work for practice (not to hand in, since the second exam takes place on Thursday, March 29).

Find the radius of convergence of the power series \[ \sum_{n=1}^\infty \frac{(n+1) i^n }{3^n} z^n.\]
Find the radius of convergence of the power series \[ \sum_{n=1}^\infty \frac{ 4+i^n}{4^n+i} z^{2n}.\] (Notice that the exponent of \(z\) is \(2n\), not \(n\). In other words, all the odd powers of \(z\) have coefficient equal to zero.)
Find the radius of convergence of the power series \[ \sum_{n=1}^\infty 2^n z^{2^n}.\] (Notice that the exponent of \(z\) is \(2^n\). This series is an example of a so-called gap series. Most of the powers of \(z\) have coefficient equal to zero, and there are progressively larger gaps between successive nonzero terms.)

Answers for the three problems: The three radii of convergence are \(3\), \(2\), and \(1\), respectively.

In reviewing for the upcoming exam, you might like to look at the second exam from last semester and the solutions.

Tuesday, March 20

Here is the assignment for next time (not to hand in, but be prepared for a quiz).

Problem 6.37, which says to show that the series \[\sum_{n=1}^\infty \frac{z^{n-1}}{2^n} \] converges when \(|z|\lt 2\) and to find the sum of the series.
Problem 6.39, which says to determine the values of \(z\) for which the series \[ \sum_{n=1}^\infty \frac{1}{(z^2+1)^n}\] converges and to find the sum of the series.
Problem 6.60, which asks for the region of absolute convergence of the series \[ \sum_{n=1}^\infty \frac{n(-1)^n(z-i)^n} {4^n (n^2+1)^{5/2}}.\]

In class, we discussed the geometric series and three convergence tests for infinite series: the comparison test, the ratio test, and the root test (all of which are tests for absolute convergence).

Thursday, March 8

There is no assignment to do over Spring Break.
In class today, we applied Liouville's theorem to prove the fundamental theorem of algebra, we derived Gauss's mean-value theorem from Cauchy's integral formula, and we discussed the maximum principle for the modulus of an analytic function. Then we took the following quiz.

Evaluate \(\displaystyle \frac{1}{2\pi i}\int\limits_{|z|=2} \frac{\sin(\pi z)}{(z-1)(z-3)^2}\,dz\).
Evaluate \(\displaystyle \frac{1}{2\pi i}\int\limits_{|z|=4} \frac{\sin(\pi z)}{(z-1)(z-3)^2}\,dz\).
Evaluate \(\displaystyle \frac{1}{2\pi}\int_0^{2\pi} \frac{e^{i\theta}+2}{3e^{i\theta}+4}\,d\theta\).

Here are solutions to the quiz.

View the integrand as \(f(z)/(z-1)\), where \(f(z)=(z-3)^{-2}\sin(\pi z)\). By Cauchy's integral formula, the answer is \(f(1)\), which simplifies to \(0\) since \(\sin(\pi)=0\).
In view of the principle we have discussed in case there is more than one singularity inside the curve, the problem can be solved by integrating around a small circle centered at \(1\), integrating around a small circle centered at \(3\), and adding the results. The first subproblem is equivalent to Problem 1, already computed as having the answer \(0\). The second subproblem can be handled by Cauchy's integral formula for the first derivative, viewing the integrand as \(g(z)/(z-3)^2\), where \(g(z)=(z-1)^{-1}\sin(\pi z)\). The answer is \(g'(3)\). You could compute the derivative of \(g\) either by the product rule or by the quotient rule. Here is the computation by the product rule (which perhaps is less prone to error): \[ g'(z) = -(z-1)^{-2}\sin(\pi z) + (z-1)^{-1}\pi\cos(\pi z). \] Then \(g'(3)=0+2^{-1}\pi\cos(3\pi) = -\pi/2\), which is the final answer to the problem.
The problem is an application of Gauss's mean-value theorem. The difficulty is to decide what function is being averaged on what circle. Different choices are possible. Perhaps the simplest choice is to use a circle of radius \(1\) centered at \(0\), and to take \(f(z)\) equal to \[\frac{z+2}{3z+4}. \] This function has a singularity at \(-4/3\), which is outside the unit circle. The problem asks for the average value of \(f(e^{i\theta})\), which equals the value of \(f(z)\) at the center of the circle: namely, \(f(0)\), which equals \(2/4\), or \(1/2\).

Tuesday, March 6

Here is the assignment for next time (not to hand in, but as usual be prepared for a quiz).

Problem 5.38(b), which says to evaluate the integral \[\oint_C \frac{(\sin(z))^6} {(z-\frac{\pi}{6})^3}\,dz \] when \(C\) is the circle of radius \(1\) centered at the origin.
Problem 5.39, which says to evaluate \[\frac{1}{2\pi i} \oint_C \frac{e^{zt}} {(z^2+1)^2}\,dz \] when \(t\gt 0\) and \(C\) is the circle of radius \(3\) centered at the origin.
Hint: You worked on a similar problem at the end of class today.
Read the statement of Gauss's mean-value theorem in Section 5.2. (In mathematics, the word “mean” means “average.”) Then do Problem 5.52, which says to evaluate \[\frac{1}{2\pi} \int_0^{2\pi} [\sin(\tfrac{\pi}{6} + 2 e^{i\theta})]^2\,d\theta. \]

In class, we discussed the homework problem about doubly periodic functions and the version of Cauchy's integral formula for derivatives. Then we worked in groups on the following two integrals: \[\oint_{|z-i|=1} \frac{e^{\pi z}}{(z^2+1)^2}\,dz \] and \[\oint_{|z|=2} \frac{e^{\pi z}}{(z^2+1)^2}\,dz. \]

Thursday, March 1

Here is the assignment to hand in next time.

Problem 5.32b, which says to evaluate the integral \[\oint_C \frac{e^{3z}}{z-\pi i}\,dz \] when \(C\) is the ellipse (understood to be oriented counterclockwise) defined by the following equation: \[ |z-2|+|z+2|=6. \] Hint: The main issue is to figure out whether the singularity lies inside the ellipse or outside.
Read Section 5.1 in the textbook, and observe that formula (5.2) is a turbocharged version of Cauchy's integral formula involving derivatives. Apply that formula to solve Problem 5.35, which says to evaluate \[ \oint_C \frac{e^{iz}}{z^3}\,dz \] when \(C\) is the circle of radius \(2\) centered at the origin.
Problem 5.49, which says to show that if \(a\) and \(b\) are specified positive real numbers, and \(F\) is a nonconstant function such that \(F(z+a)=F(z)\) for every complex number \(z\) and \(F(z+bi)=F(z)\) for every complex number \(z\), then \(F\) cannot be analytic throughout the rectangle where \(0\le x\le a\) and \(0\le y\le b\). (In other words, the function \(F\) must have at least one singularity somewhere in the rectangle.)
Hint: This problem is in the section titled “Liouville's theorem.”
Remark: The usual formulation of the statement is that constant functions are the only doubly periodic entire functions.

In class, we proved Cauchy's integral formula and Liouville's theorem. Then we took the following quiz.

State (a) Cauchy's integral theorem and (b) Cauchy's integral formula.
Evaluate \[\int_C \frac{3z^2 + \cos(\pi z)}{z-1} \,dz,\] where \(C\) is the circle centered at \(0\) of radius \(2\) (oriented in the usual counterclockwise direction).
Give an example of a simple closed curve \(C\) for which \[ \int_C \frac{1}{z(z-2)}\,dz = \pi i.\]

Here are solutions.

Cauchy's integral theorem says that if a function \(f\) is analytic on and inside a simple closed curve \(C\), then \(\int_C f(z)\,dz=0\). Cauchy's integral formula says that if additionally \(z_0\) is a point of the region inside the curve \(C\), then \[\int_C \frac{f(z)}{z-z_0}\,dz = 2\pi i\times f(z_0).\]
Apply Cauchy's integral formula with \(z_0\) equal to \(1\). The value of the indicated integral is \(2\pi i\times (3+\cos(\pi))\), which simplifies to \(4\pi i\).
We previously computed this integral when the integration path \(C\) is the circle of radius \(1\) centered at \(0\), oriented in the standard counterclockwise direction, and got the answer \(-\pi i\). One way to get the answer \(+\pi i\) is to use the same integration path but oriented in the clockwise direction, which reverses the sign of the integral. An alternate way to get the answer \(+\pi i\) is to take \(C\) to be a circle of radius \(1\) centered at \(2\) and oriented counterclockwise. Other answers are possible. But what will not do is to integrate along a path that encircles both singularities or neither singularity; in those cases, the integral equals \(0\).

Tuesday, February 28

Here is the assignment for next time (not to hand in, but be prepared for a quiz).

Problem 5.30, which says to evaluate \[ \frac{1}{2\pi i} \oint_C \frac{e^z}{z-2}\,dz \] when the integration path \(C\) is (a) the circle of radius \(3\) centered at the origin and (b) the circle of radius \(1\) centered at the origin.
Problem 5.79, which says to evaluate \[ \frac{1}{2\pi i} \oint_C \frac{z^2}{z^2+4}\,dz \] when the integration path \(C\) is a square with vertices at \(\pm 2\) and \(\pm2+4i\). (The path is understood to be traversed in the usual counterclockwise direction.)
Hint: You know how to factor the denominator.
Problem 4.102, which says to show that if \(n\) is a positive integer, then \[ \int_0^{2\pi} e^{\sin(n\theta)} \cos(\theta-\cos(n\theta))\,d\theta=0 \] and \[ \int_0^{2\pi} e^{\sin(n\theta)} \sin(\theta-\cos(n\theta))\,d\theta=0. \] Hint: It looks hopeless to compute these integrals via real calculus methods. But an integral from \(0\) to \(2\pi\) can arise by parametrizing a complex line integral taken over the unit circle, setting \(z\) equal to \(e^{i\theta}\). Could the given integrals be the real part and the imaginary part of some complex line integral that equals \(0\) by Cauchy's integral theorem?

In class, we discussed the complex-variables version of the fundamental theorem of calculus, Cauchy's integral formula, and Liouville's theorem.

Thursday, February 23

Here is the assignment to hand in next time.

Problem 4.35, which says to evaluate \(\oint_C |z|^2\,dz\) when \(C\) is the boundary of the unit square. The integration path goes from the point \( (0,0)\) to \( (1,0)\) to \((1,1)\) to \((0,1)\) and back to \((0,0)\).
Warning: The integrand is not an analytic function!
Problem 4.53, which says to show that if \(C\) is a simple closed curve, then \[\frac{1}{2i} \oint_C \overline{z}\,dz \] equals the area enclosed by the curve.
Notice that the integrand is not an analytic function, so Cauchy's integral theorem does not apply. But Green's theorem does apply.
Problem 4.93, which says to show that \[\int_0^\infty xe^{-x}\sin(x)\,dx=\frac{1}{2}. \] Hint: In principle, you could work this integral as a real calculus integral via multiple applications of integration by parts. But there is a short-cut. By Euler's formula, you can rewrite \(\sin(x)\) as \(\Im (e^{ix})\), so the problem turns into \[\Im\int_0^\infty x e^{-x+ix}\,dx. \] Now a single application of integration by parts suffices.
Problem 5.81a, which says to show that \[\oint_C \frac{1}{z+1}\,dz=2\pi i \] when \(C\) is the circle of radius \(2\) centered at the origin.
Although we have not yet started Chapter 5, you should be able to solve this problem by using the principles that we discussed in class today.

In class, we discussed Cauchy's integral theorem and the principles of path independence and path deformation for line integrals of analytic functions. We also computed the following important integral by parametrizing the path: \[\int_{|z|=1} \frac{1}{z}\,dz=2\pi i. \]

Tuesday, February 21

Here is the assignment to hand in next class.
[Notice that there are answers in the book. The main goal of this assignment is to refresh your memory about some integration techniques that you encountered in your calculus courses.]

Problem 4.38a, which says to evaluate the line integral \[ \int_C (3xy+iy^2)\,dz \] when \(C\) is the straight line segment joining the point \(i\) to the point \(2-i\).
[To get started, you need to find a parametrization of the line segment.]
Problem 4.46, which says to evaluate the line integral \[ \oint_C (5x+6y-3)\,dx + (3x-4y+2)\,dy \] around a triangle in the \(xy\) plane with vertices at \( (0,0)\), \((4,0)\), and \((4,3)\).
Hint: Apply Green's theorem. The symbol \(\oint\) signifies a line integral along a closed curve.
Problem 4.79b, which says to evaluate the indefinite integral (antiderivative) \[ \int z \sin(z^2)\,dz \] (think substitution).
Problem 4.80a, which says to evaluate the indefinite integral \[ \int z\cos(2z)\,dz \] (think integration by parts).

In class today, I returned the graded exams, and we reviewed line integrals and Green's theorem from vector calculus. We concluded by applying Green's theorem to deduce that \(\int_C f(z)\,dz=0\) when \(f\) is an analytic function in the region bounded by the simple closed curve \(C\).

Friday, February 17

I have posted the first exam and solutions.

Thursday, February 16

The first exam was given. I will post solutions tomorrow.
Next time we will start Chapter 4. The assignment for over the weekend is to find out what famous complex analyst died one hundred and fifteen years ago on February 19.

Tuesday, February 14

In class today, we came to a consensus on highlights from Chapters 1–3, and we solved some sample exam problems in groups for a quiz grade.

By the way, solutions are available for the first exam from last semester.

The assignment, of course, is to study for the exam to be given on Thursday.

Thursday, February 9

The assignment is to start reviewing for the exam; and

to list two highlights for each of Chapters 1, 2, and 3 (highlights are items such as fundamental concepts, important theorems, and key formulas); and
to compose a sample exam problem for each of Chapters 1, 2, and 3 (meaning a problem that addresses an important topic and that other students in the class should be able to solve in a reasonable amount of time).

In class today, we discussed various techniques for solving the three problems assigned last time: namely, l'Hôpital's rule, expansion in infinite series, the method of undetermined coefficients, and the Cauchy–Riemann equations. The exam from last semester that I displayed at the end of class is available online. (Problem 6 on that exam is a Chapter 4 topic that we have not yet covered.)

Tuesday, February 7

Here is the assignment for next time (not to hand in, but be prepared for a quiz).

Problem 3.81, which says to evaluate \[\lim_{z\to 0} \left( \frac{\sin(z)}{z}\right)^{1/z^2}. \] Notice that the complex power implicitly involves a logarithm. There is an unstated assumption that the branch of the logarithm function is chosen such that \(\log(1)\) equals \(0\) (rather than \(2\pi i\) or some other multiple of \(2\pi i\)). The answer is in the book.
Problem 3.104, which says to find an analytic function \(f(z)\) such that the real part of the derivative \(f'(z)\) equals \(3x^2-4y-3y^2\), and \(f(1+i)=0\).
This problem is related to the problem you did last time about finding a harmonic conjugate. How is the real part of the derivative \(f'(z)\) related to the partial derivatives of \(u\) and \(v\)? The value of \(f\) at \(1+i\) enters as a boundary condition that enables you to evaluate integration constants.
The answer in the book is correct but is not the most general possible answer, which is \(z^3+2iz^2+6-2i + a(iz+1-i)\), where \(a\) denotes an arbitrary real constant.
Problem 3.111, which says to show that if \(u\) and \(v\) are two harmonic functions (but not necessarily the real part and the imaginary part of the same analytic function), then the following function is analytic: \[\left( \frac{\partial u}{\partial y} - \frac{\partial v}{\partial x}\right) + i \left( \frac{\partial u}{\partial x} +\frac{\partial v}{\partial y}\right). \]

In class, we discussed Laplace's equation, harmonic functions, and the notion of conformality.

Thursday, February 2

Here is the assignment to hand in next time.

Problem 3.97, which says to find an equation for the line normal to the curve \(x^2y=2xy+6\) at the point \( (3,2)\).
[This problem is about real calculus; the problem is connected with the discussion in class today about gradient vectors and orthogonal curves.]
Suppose \(u(x,y)=y-e^x\) and \(v(x,y)=y-e^{-x}\). Show that the level curves of \(u\) are orthogonal to the level curves of \(v\), yet the function \(u+iv\) is nowhere analytic.
Read Section 3.4 in the textbook (about harmonic functions). Then do Problem 3.53(a), which says to show that if \(u(x,y)=3x^2y+2x^2-y^3-2y^2\), then \(u\) is harmonic; moreover, find a function \(v\) such that \(u+iv\) is analytic, and express \(u+iv\) as a function of \(z\). (You can use the solutions in the textbook to Problems 3.7 and 3.8 as models.)

In class, we studied the geometry of the mapping that sends the complex variable \(z\) to \(z^2\). We found the level curves of \(\Re(z^2)\) and \(\Im(z^2)\), and we found the image of the unit square.
Next time, I plan to discuss the concept of conformality and the classification of singular points.
The following quiz was given in class.

State the Cauchy–Riemann equations for a function \(f=u+iv\).
Give an example of a function that is analytic in the entire complex plane.
Give an example of a function that is nowhere analytic.

Solutions

The Cauchy–Riemann equations say that \(\dfrac{\partial u}{\partial x}=\dfrac{\partial v}{\partial y}\) and \(\dfrac{\partial u}{\partial y} = - \dfrac{\partial v}{\partial x}\). The alternate complex form says that \(\dfrac{\partial f}{\partial x} = -i\dfrac{\partial f}{\partial y}\).
Every polynomial in \(z\) is analytic in the entire complex plane. In particular, if \(f(z)=z^2\), then \(f\) is an entire function.
If \(f(z)=\overline{z}\) (the complex conjugate), then \(f\) is nowhere analytic. We did this example in class last time. There was another example in the homework for today: the function \(x^2+iy^3\) is nowhere analytic.

Tuesday, January 31

In class, we discussed the notion of complex differentiability and the Cauchy–Riemann equations. Here are some exercises for practice. These exercises are not to be turned in, but be prepared for a quiz in class on Thursday.

Problem 3.46, which asks to find the derivative at points where the function is analytic and also to determine the singular points (the points where the function is not analytic) for (a) \(\dfrac{z}{z+i}\) and (b) \(\dfrac{3z-2}{z^2+2z+5}\).
Problem 3.47(b), which says to deduce that \(ze^{-z}\) is an analytic function by verifying the Cauchy–Riemann equations.
Problem 3.48, which says to show that the function \(x^2+iy^3\) is not analytic anywhere, yet the Cauchy–Riemann equations are satisfied at the origin (the point where \(x=0\) and \(y=0\)).
[There is a technical point here: namely, the word analytic refers to a property that holds on an open set.]

Thursday, January 26

Here is the assignment to hand in next time.

Problem 2.61(b), which says to find all values of \(z\) for which \(e^{4z}=i\).
Problem 2.98, which says to show that \[\lim_{z\to 1} \frac{\sqrt{z^2+3}-2}{z-1} =\frac{1}{2}, \] where the branch of the square root function is chosen to be positive on the positive part of the real axis. (In other words, \(\sqrt{4}\) is supposed to be \(+2\), not \(-2\).)
Problem 2.104, which says to show that the function \(z/(z^4+1)\) is continuous except at four points; what are the four exceptional values of \(z\)?
Problem 2.135, which says to show that \[\lim_{z\to \infty} \frac{z^3-3z+2} {z^4+z^2-3z+5}=0. \]

In class today, we discussed the notions of limit and continuity for functions of a complex variable. Then we took the following quiz, for which solutions are given below.

Find all three cube roots of \(8i\). Write the answers in \(a+bi\) form.
Which quantity has bigger real part, \(\log(i+2)\) or \(\log(i-2)\)?
All of the values of the expression \(2^i\) lie on the same line in the complex plane. What is the slope of that line?

In each problem, you must show work or give an explanation.

Solutions

Since \(8i=8\exp(i\frac{\pi}{2}+2\pi n i)\) for an arbitrary integer \(n\), the cube roots of \(8i\) have the form \(2\exp(i\frac{\pi}{6} + i \frac{2\pi n}{3})\). Three different answers arise as \(n\) varies over the integers: when \(n=0\), the value of the cube root is \(2\exp(i\frac{\pi}{6})\), which simplifies by Euler's formula to \(\sqrt{3}+i\); when \(n=1\), the value is \(2\exp(i\frac{5\pi}{6})\), which simplifies to \(-\sqrt{3}+i\); and when \(n=2\), the value is \(2\exp(i\frac{3\pi}{2})\), which simplifies to \(-2i\).
The real part of \(\log(i+2)\) is \(\ln|i+2|\) or \(\ln\sqrt{5}\), and the real part of \(\log(i-2)\) is \(\ln|i-2|\) or \(\ln\sqrt{5}\). The two values are equal. Even without calculating, you should be able to see geometrically that \(|i+2|=|i-2|\), for the complex numbers \(i+2\) and \(i-2\) are reflections of each other across the imaginary axis.
The values of \(2^i\) have the form \(\exp[i(\ln 2 + 2\pi ni)]\), where \(n\) is an arbitrary integer; equivalently, \(e^{-2\pi n} e^{i\ln 2}\). These values all lie in the first quadrant on a ray with angle \(\ln 2\). The corresponding slope is \(\tan(\ln 2)\).

Tuesday, January 24

We discussed limits of sequences of complex numbers, the complex logarithm function, and complex powers. Here are some exercises for practice. These exercises are not to be turned in, but be prepared for a quiz in class on Thursday.

Limits:

\(\displaystyle\lim_{n\to \infty} \frac{n^2 i^n}{n^3+1}\) (Problem 2.117a)
\(\displaystyle\lim_{n\to \infty} n\left( \frac{1+i}{2}\right)^n\) (Problem 2.119)
\(\displaystyle\lim_{n\to \infty} \left( \sqrt{n+2i} -\sqrt{n+i}\,\right)\)
(What is the trick for simplifying a difference of square roots? This exercise is Problem 2.123c.)

Complex powers:

\((-1)^{1/4}\)
(We computed this quantity in class last time. Does the new formula \(z^w = e^{w \log(z)}\) give the same four answers?)
\(i^i\)
(Remarkably, all the values of this expression are real numbers! See Problem 2.18c.)
\(\Re[(1-i)^{1+i}]\)
(This expression is Problem 2.82a. There is an answer in the book on the last page of Chapter 2. Another form of the answer is \(e^{\frac{\pi}{4}+2n\pi}\left(\cos(\tfrac{1}{2}\ln2) + \sin(\tfrac{1}{2}\ln2) \right)\).)

Thursday, January 19

In class, we discussed the power series expansions of the complex exponential and trigonometric functions, Euler's formula (\(e^{i\theta}=\cos\theta+i\sin\theta\)), and the computation of roots of complex numbers.
The homework to turn in at the beginning of next class is the following set of four exercises (which we started working on in groups at the end of class).

1.96a: Find the cube roots of \(8\) (three answers: \(2\), \(-1+\sqrt{3}\,i\), and \(-1-\sqrt{3}\,i\)).
1.95e: Find \(64^{1/6}\) (six answers).
1.95f: Find \(i^{2/3}\) (three answers).
1.102: Solve the equation \(z^4+z^2+1=0\) (four answers).

Tuesday, January 17

The first class meeting introduced the basic notions and terminology of the field of complex numbers.
The homework to turn in at the beginning of next class is the following set of four exercises (which we started working on in groups at the end of class).

Describe geometrically the set of points \(z\) in the complex plane satisfying each of the following conditions.

\(\Re(iz)=2\)
\(|z-i|=2\)
\(z+\overline{z}=2\)
\(\left| \dfrac{z-1}{z+1}\right|=2\)

You should find that two of the equations represent lines, and the other two equations represent circles (and you should determine the center and the radius of each circle).

Monday, January 9, 2012

This site went live today. Welcome to Math 407. The first class meeting is Tuesday, January 17.