Proof of complex number identity

Click For Summary
SUMMARY

The forum discussion focuses on proving the complex number identity using the argument properties of complex numbers, specifically the identity arg(zw) = arg(z) + arg(w). The main goal is to demonstrate that arg(z^n) = n * arg(z) through mathematical induction. The discussion emphasizes the importance of establishing the base case for z^1 and then proving the inductive step for n = k, leading to n = k + 1. The second part of the question builds on this proof, utilizing the established relationship.

PREREQUISITES
  • Understanding of complex numbers and their properties
  • Familiarity with mathematical induction
  • Knowledge of de Moivre's theorem
  • Basic skills in manipulating complex arguments
NEXT STEPS
  • Study the proof of de Moivre's theorem in detail
  • Learn about mathematical induction techniques in depth
  • Explore the properties of complex arguments and their applications
  • Practice problems involving complex number identities and proofs
USEFUL FOR

Students studying complex analysis, mathematicians interested in number theory, and educators teaching advanced algebra concepts.

Gameowner
Messages
43
Reaction score
0

Homework Statement



Attached question

Homework Equations





The Attempt at a Solution



The second part of question is relatively easy, it is the first part of the question where I need help with(using arg zw = arg z + arg w to show arg z^n = n arg z).

Also, is the question asking to proof de Moivre's formula with the identity arg zw = arg z + arg w? or just proof LHS=RHS?
 

Attachments

  • pic.JPG
    pic.JPG
    15 KB · Views: 526
Physics news on Phys.org
I guess it's the induction that's causing problems. Show that it's true for z2. Then show that if it's true for zn, that it's also true for zn+1.

The second part uses the relation you proved in the first part.
 
The base case would be for z1, I thought. In any event, it's easy to prove.
The inductive case: assume true for n = k, show that it's true for n = k + 1.
Here's a hint: xa + 1= (xa)(x).69
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 29 ·
Replies
29
Views
4K
  • · Replies 6 ·
Replies
6
Views
6K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 35 ·
2
Replies
35
Views
11K
  • · Replies 9 ·
Replies
9
Views
6K