Question about complex numbers

Click For Summary

Discussion Overview

The discussion revolves around calculating the expression (\frac{\sqrt{3}+i}{2})^{2010} without employing De Moivre's theorem. Participants explore various methods and express their thoughts on the feasibility of the task.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant suggests using binomial expansion as a potential method for calculation.
  • Another participant humorously expresses frustration at the complexity of the problem without De Moivre's theorem.
  • A suggestion is made to use Wolfram Alpha for computation, indicating that manual calculation would be cumbersome.
  • A participant discusses breaking down the expression to eliminate the imaginary part, referencing their approach with (1+i)^{2010} as a comparison.
  • One participant mentions that raising the number to the 12th power results in 1, proposing a method to prove this by hand and questioning the avoidance of De Moivre's theorem.
  • Another participant reiterates the idea of using the binomial theorem to simplify the calculation by raising the expression to the third power.

Areas of Agreement / Disagreement

Participants express differing opinions on the best approach to solve the problem, with no consensus on a single method. Some advocate for binomial expansion while others highlight the challenges of avoiding De Moivre's theorem.

Contextual Notes

Participants acknowledge the complexity of the problem and the potential limitations of their proposed methods, particularly in relation to the imaginary components of the expression.

Chuckster
Messages
20
Reaction score
0
Hello guys!

I have a question related to complex numbers.

How would i calculate, for example

[tex](\frac{\sqrt{3}+i}{2})^{2010}[/tex] without using the De Moivre's theorem?
 
Physics news on Phys.org
Binomial expansion?
 
Curl up in a ball and die a little inside.
 
Use wolfram|alpha to do it for you.

Really, it would be really cumbersome to do it without De Moivre's theorem.
 
SprucerMoose said:
Binomial expansion?

well, that was my idea originally.
using the facts that
[tex] i^{1}=i, i^{2}=-1, i^{3}=-i, i^{4}=1[/tex]

i tried to find a way to brake the expression given in the first post into something which could destroy the [te]i[/tex], just like i would do with, ie
[tex](1+i)^{2010}=(2i)^{1005}=2^{1005}i[/tex], but I'm having problems shaking off the real part of the imaginary number.
 
Using De Moivre's theorem we can immediately observe that if we raise your number to the 12th power you get 1. So prove this with pencil and paper by finding the twelfth power by hand, seeing that it's one, and then you're basically done. You can do this in half the work by just raising it to the sixth power and noticing that you get negative the starting number.

The question then is why are you trying to avoid De Moivre's theorem: because you want avoid using it, or because somebody else is forbidding you from using it in your final solution
 
Office_Shredder said:
You can do this in half the work by just raising it to the sixth power and noticing that you get negative the starting number.
You can even raise it to the 3rd power using the Binomial theorem and see what you get.
 

Similar threads

  • · Replies 108 ·
4
Replies
108
Views
13K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 13 ·
Replies
13
Views
6K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K