Mathematical Methods II (2014 edition)

This page gathers all the material of the Mathematical Methods II that is specific to the 2014 edition.


20.01: Introducing Complex Numbers · handout 1.
21.01: Complex functions of complex numbers · handout 2.
22.01: Exponentials, trigonometric functions, hyperbolics and their inverses · handout 3.
27.01: Limits and continuity (for the Physicist) · handout 5.
28.01: Limits and continuity (for the Mathematician) · handout 6.
3.02: Derivatives and analyticity · handout 7.
4.02: Differentiability and Cauchy-Riemann · handout 8.
10.02: Harmonic functions and Laplace's equation · handout 9.
11.02: Complex Potentials · handout 10.
17.02: Conformal mapping · handout 11.
18.02: The Möbius Transformation · handout 12.
24.02: Integrals in the complex plane · handout 13.
25.02: Line and contour integrals · handout 14.
3.03: The Cauchy-Goursat theorem and its integral forms · handout 15.
4.03: Consequences of the Cauchy theorem · handout 16.
10.03: Series of Complex Numbers · handout 17.
11.03: Power Series · handout 18.
17.03: Uniform convergence · handout 19.
18.03: Analytic functions and continuation · (no handout).
24.03: Taylor Series · handout 20.
25.03: Laurent Series · handout 21.
31.03: Singularities · handout 22.
01.04: Residues · handout 23.
7.04: Applications of Residue Theory · handout 24.
8.04: Fourier Series · handout 25.
22.04: Summation of Series · handout 26.
23.04: The Riemann and the Bloch Spheres · handout 27.
28.04: Solutions of home exams I.
29.04: Solutions of home exams II.
30.04: All the questions answered.

Continuous Evaluation

The final note was composed of 30% continuous evaluation, 35% partial exam and 35% the final exam.

Dates by which to return the home exam:

  1. 30 January in Handout 3.
  2. 13 February in Handouts 7 & 8.
  3. 3rd of March in Handout 12.
  4. 13 March (in class)
  5. 7 April in Handout 20.
  6. 22 April in Handout 24.


Solutions of the Continuous Evaluation:

  • Solution of #1 in Handout 10.
  • Solution of #2 in Handouts 15 & 16.
  • Solution of #3 in class.
  • Solution of #4 in Handout 19.
  • Solution of #5 and #6 in this supplementary material.

Final Examination

Was held on 16th of May (Friday), from 15.00 till 18.00, in the rooms 101.4 and 101.6.

There have been 60 persons presenting, 58 works returned. 37 (64%) achieve a mark higher than 5 (pass). The highest note is 10 (full score), the smallest is 1, the average is 5.6.

The details of the notes appear in this file. First four numerical columns are the notes broken down over the corresponding exercises (each noted on five). The last column is the note that results.

Final Note

The final note takes into account for 35% the continuous examination to help students who performed well before the final exam. It does not lower the note. It is computed as follows:

  • The partial exam counts for 75% of the continuous evaluation note.
  • The home exam count for 25%. Best of two works is retained. If only one, or no, work was returned, the partial counts for 100% of the continuous evaluation. If the home exams lower the note, also only the partial counts.

Recovery Exam / Examen de Recuperación

23 June 2014 01.00.AU.206 - AULA 206 MÓDULO 00

Conditions are the same as for the final exam, allowing the use of notes and books.

The questions (three hours).


Question 1



The poles are easily found by noticing that $(1+x^2+x^4)(1-x^2)=1-x^6$, i.e., they are the sixth root of unity which are not solutions of $x^2=1$, which leaves:

\begin{equation} e^{ik\pi/3}\quad\mathrm{for\ }k\in\{1,2,4,5\} \end{equation}

We can now calculate the integral by computing the residues and summing those on the upper half-plane (times $2i\pi$):

\begin{multline} \mathrm{Res}_{k=1}\frac{1}{(z-e^{i\pi/3})(z-e^{i2\pi/3})(z-e^{-i\pi/3})(z-e^{-i2\pi/3})} =\\ \frac{1}{(e^{i\pi/3}-e^{i2\pi/3})(e^{i\pi/3}-e^{-i\pi/3})(e^{i\pi/3}-e^{-i2\pi/3})}\,. \end{multline}

The denominator is calculated easily by dealing with the geometry of the unit circle (one is a sine, the other is after computing a product, the translated sum of a number with its opposite). One thus finds the residue as:


The other residue is found similarly as


so that the integral is finally, since $1/e^{-i\pi/3}+1/e^{i\pi/3}=e^{-i\pi/3}+e^{i\pi/3}=1$, simply:


Question 2

We apply the Cauchy-Riemann criterion:

\begin{align} \partial_x\cos x&=-\sin x&\, \partial_x(-\sinh y)&=0\\ \partial_y\cos x&=0&\, \partial_y(-\sinh y)&=-\cosh y \end{align}

While one condition ($\partial_x v=-\partial_y u$) is always satisfied, the other ($\partial_x u=\partial_y v$) is only for points where both the cosine and the hyperbolic sine are unity, since one is less than one and the other larger than one otherwise. The function is therefore derivable at points $\pi/2+2k\pi$ for $k\in\mathbf{Z}$ on the real line.

Question 3

A function $f$ is harmonic of $\nabla^2 f=0$, i.e., $(\partial_x^2+\partial_y^2)f=0$, which in our case translates as:

\begin{equation} n(n-1)(x^{n-2}-y^{n-2})=0 \end{equation}

for all $x$ and $y\in\mathbf{R}$. This is achieved when $n=0$, and 1 by direct cancellation of the prefix, but also for $n=2$ by cancellation of the function itself $x^0=y^0=1$. For other integer values of $n$, the Laplacian is a nonzero function of $x$ and $y$ and the function therefore not harmonic.

Question 4

We write $\sin(z)=\sin(x+iy)$ with $x$ and $y$ real, and expand trigonometrically:

\begin{equation} \sin(x+iy)=\sin x\cosh y+i\cos x\sinh y \end{equation}

so that

\begin{align} |\sin(x+iy)|^2&=\sin^2x\cosh^2y+\cos^2x\sinh^2y\\ &=\sin^2x\cosh^2y+\sin^2x\sinh^2y-\sin^2x\sinh^2y+\cos^2x\sinh^2y\\ &=\sin^2x+\sinh^2y \end{align}

here we have added and removed $\sin^2x\sinh^2y$ and used $\sin^2+\cos^2=1$ and $\cosh^2-\sinh^2=1$.

Now $\sinh(y)$ is zero only for $y=0$ (this is the only solution to $e^y=e^{-y}$), therefore all the zeros of $\sin(z)$ are real, and are given by the zeros of $\sin(x)$, i.e., $z=\pi\mathbf{Z}$.

It could also be done as seen in class, reducing the problem to the solutions of $e^{2iz}=1$.


The final note is calculated as given by the Guía Docente, which is 45% exam + 25% partial + 30% homework. While I believe giving over 50% of the weight to work done at home does not best reflect the level, three students have complained about my using a stronger weighting for the class exam, and their wish has been followed blindly and indiscriminately by the competent instances. This made a few people pass without the slightest idea of what is a pole, a derivative in the complex plane or related basic notions of complex analysis. While I lament this state of things, such is the policy of the Universidad Autónoma de Madrid, to which I leave the responsibility of the level of the students that it qualifies. Interestingly, even in such conditions, some people manage to fail.

See Also