Recent content by Advent

  1. Advent

    MHB Closure of a Subgroup of GL(2,C)

    I understand your proof and the closure of \bar{G} follows quite easily just by multiplying the matrices, however I'm not quite convinced on the solution of this exercise. I'm quite grateful on your help, but I think I need to study some prerequisites of this book and in mathematics in general...
  2. Advent

    MHB Closure of a Subgroup of GL(2,C)

    I will think about your post and reply here later, thank you very much.
  3. Advent

    MHB Closure of a Subgroup of GL(2,C)

    So you use that the closure of a set is itself plus its boundary? Let me explain it myself to see if I'm understanding this: Because G \subseteq \bar{G} we only need to compute the boundary part. Now we want to see if the number \frac{a(t+2\pi n)}{s+2\pi m } << 1 but I can't put that expression...
  4. Advent

    MHB Closure of a Subgroup of GL(2,C)

    Let GL(2;\mathbb{C}) be the complex 2x2 invertible matrices group. Let a be an irrational number and G be the following subgroup G=\Big\{ \begin{pmatrix}e^{it} & 0 \\ 0 & e^{iat} \end{pmatrix} \Big| t \in \mathbb{R} \Big\} I have to show that the closure of the set G is\bar{G}=\Big\{...
  5. Advent

    MHB Rank & Letter of $\bf{\mathbb{INDONESIA}}$ in Dictionary

    You may find this link interesting Surya's Blog!: Finding Rank of a Word: With Repetition
  6. Advent

    Other Should I Become a Mathematician?

    I've only read half topic but it has an insane amount of advice, references, and enjoyable stuff. Thank you all, seriously.
  7. Advent

    Triangle inequality for complex numbers: sketch of proof

    Homework Statement Show that if z_1,z_2 \in \mathbb{C} then |z_1+z_2| \leq |z_1| + |z_2| Homework Equations Above. The Attempt at a Solution I tried by explicit calculation, with obvious notation for a,b and c: my frist claim is not that the triangle inequality holds, just that...
  8. Advent

    MHB Complex Integral: Solving a Difficult Problem

    To be sure if I understood this, my answer to the first integral is $i\pi^3/4$. If $c=\{2+e^{i\theta} : \theta \in [0,2\pi] \}$ as the poles are all in the imaginary axis, and $c$ is the circle of radius $1$ and center $2$ it never touches the imaginary axis, therefore no poles inside $c$, so...
  9. Advent

    MHB Mathematics Study: How to Master the Theory

    Hi! I'm wondering how do we study mathematics. If the book has exercises, after reading (maybe several times) the theory, you can go and do the exercises. If the book is only theory and no exercises, how do you check your understanding on the subject?
  10. Advent

    MHB Complex Integral: Solving a Difficult Problem

    Thank you, now I get it! I didn't know the $\varepsilon(z)$ thing, my math training is the one I am being given in my physics undergraduate courses so it's quite mechanical and we don't care too much about math rigor. Such a bad mistake in my opinion, but I'm my spare time, wich is not as much...
  11. Advent

    MHB Complex Integral: Solving a Difficult Problem

    Hi, thanks for the hint! Ok, now I see it, after some google research, that your result for the pole is the same as L'Hôpital's rule $$\lim_{z\to p}(z-p)\frac{g(z)}{h(z)}=\lim_{z \to p}\frac{g(z)}{\frac{h(z)}{z-p}}=\lim_{z \to p}\frac{g(z)}{\frac{h(z)-h(p)}{z-p}}=\lim_{z \to...
  12. Advent

    MHB Complex Integral: Solving a Difficult Problem

    Hi, thanks for your reply. The residue is: $$\lim_{z\to i\pi/2}\left(z-\frac{i\pi}{2}\right)\frac{z^2}{e^{2z}+1}$$ Now, the "problem" is that I don't know to "factorize" $e^{2z}+1=0$, to cancel with $\left(z-\frac{i\pi}{2}\right)$ I was thinking in circular and hyperbolical functions, but...
  13. Advent

    MHB Complex Integral: Solving a Difficult Problem

    Hi all! I have to perform this complex integration over three curves, the first one is \( C=\{ z \in \mathbb{C} : |z|=2 \} \) and the function to integrate is $$ f(z)=\frac{z^2}{e^{2z}+1}$$ If I do the usual change of variables \(z=2e^{i\theta} \) and integrate from \( \theta = 0 \rightarrow...
  14. Advent

    Exploring the Infinite Speed of Time in Low Gravity Fields

    The right to think about, I guess, it's that when you are here or here, you measure some time. Every observer has some rules and some clocks to measure time. Now suppose the twin paradox. The thing that actually happens is that their clocks will not measure the same amount of time, but time...
  15. Advent

    Wave equation and Galilean Transformation

    I know, in fact i should do it for my speclal relativity exam, but there is some little work and theory before Lorentz Transformations.
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