De moivre's theorem complex number

kelvin macks
Messages
60
Reaction score
0
can anyone explain how ro make the working above the red circle to the working in the red circle? why the author do this?
 

Attachments

  • DSC_0015~3.jpg
    DSC_0015~3.jpg
    31.1 KB · Views: 908
Mathematics news on Phys.org
It's equivalent to having

(a+b)c = ac+bc

where

a=z^2+\frac{1}{z^2}

b=2

c=z^2-\frac{1}{z^2}

and why he did it should be pretty evident from his next two lines.
 
  • Like
Likes 1 person
It has nothing to do with "complex numbers" or "DeMoivre's Theorem". It is, as mentallic said, just the distributive law.
 
  • Like
Likes 1 person
Hahaha, this is something Feynman talked about from his childhood. What you've presented is actually a higher-order form of something called Morrie's[/PLAIN] Law (Feynman's little friend in childhood). From what I've studied, a useful application is in the proof of http://2000clicks.com/MathHelp/GeometryTriangleUrquhartsTheorem.aspx.
 
Last edited by a moderator:
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
I'm interested to know whether the equation $$1 = 2 - \frac{1}{2 - \frac{1}{2 - \cdots}}$$ is true or not. It can be shown easily that if the continued fraction converges, it cannot converge to anything else than 1. It seems that if the continued fraction converges, the convergence is very slow. The apparent slowness of the convergence makes it difficult to estimate the presence of true convergence numerically. At the moment I don't know whether this converges or not.

Similar threads

3
Replies
108
Views
10K
Replies
7
Views
740
Replies
2
Views
1K
Replies
3
Views
1K
Replies
8
Views
8K
Replies
6
Views
2K
Replies
6
Views
1K
Back
Top