Proving the Angle Relationship in Complex Vector Multiplication

  • Thread starter Thread starter nucleawasta
  • Start date Start date
  • Tags Tags
    Complex Vector
Click For Summary

Homework Help Overview

The discussion revolves around proving relationships in complex vector multiplication, specifically focusing on the lengths and angles of vectors defined in the complex plane. The original poster presents a problem from a textbook concerning the multiplication of two complex vectors and their geometric interpretations.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to show that the angle between the product of two complex vectors is the sum of their individual angles. Some participants question the use of small angle approximations and suggest finding a relevant trigonometric identity. Others introduce the possibility of using Euler's formula for a different approach.

Discussion Status

The discussion is active, with participants exploring different methods to address the problem. Some guidance has been offered regarding trigonometric identities, and there is acknowledgment of the potential utility of Euler's formula. The original poster has made progress by relating their findings to known trigonometric relationships.

Contextual Notes

Participants note the requirement to prove the relationships for arbitrary angles, avoiding approximations that might simplify the problem. There is also mention of the context of the problem being part of a physics curriculum, which may impose certain constraints on the methods used.

nucleawasta
Messages
5
Reaction score
0

Homework Statement



Question from Vibrations and Waves by A.P. French Chapter 1

Consider a vector z defined by Z=Z1Z2, where Z1=a+jb, Z2=c+jd.

a)Show that the length of the of z is the product of the lengths of Z1 and Z2.

b)Show that the angle between z and the x-axis is the sum of of the angles made by Z1 and Z2

Homework Equations


tan(θ1)=b/a
tan(θ2)=d/c
|Z1|=Z1
|Z2|=Z2

The Attempt at a Solution



So the first part I didn't have any trouble with, it was fairly straight forward showing that the length of Z1*Z2 was equal to the length of Z. But when I moved to part B I ran into a problem. Here's what I tried.

I Knew θ1=b/a and θ2=d/c by a first order taylor expansion of the tangents of these angles and since I am told the angle of Z, θZ is the sum of these two. I must prove:

θZ=(cb+da)/ca

However when I write out the form of Z=Z1Z2 I get:

Z=ac-bd +j(ad+bc). Now since I know the tan(θZ)=imaginary/real

I get tan(θZ)=(ad+bc)/(ac-bd).

I'm not quite sure what I'm doing wrong, but I'd really appreciate a hand! Thanks!
 
Physics news on Phys.org
You don't want to make any small angle approximation since you want to prove it for arbitrary angles.
nucleawasta said:
tan(θZ)=(ad+bc)/(ac-bd).
Try finding a trig identity involving the tangent function that you can relate to your expression for tan(θZ).
 
Have you been introduced to Euler's formula yet or are you required to solve it in cartesian form?
 
I mean I'm actually a senior physics major :P(slightly embarrassing I couldn't solve this) I've seen Euler's identity and it is introduced in the chapter, so I suppose that could be a viable way to solve the problem.
 
TSny said:
You don't want to make any small angle approximation since you want to prove it for arbitrary angles.

Try finding a trig identity involving the tangent function that you can relate to your expression for tan(θZ).


Many thanks,

Using the relation tan(θ12)=(tan(θ1)+tan(θ2))/1-tan(θ1)*tan(θ2)

I was able to use trigonometry(SOHCAHTOA as i learned it way back when) to plug in for the tan(θ1) and tan(θ2) which ultimately leads to the solution I was trying to prove from my first post.

:smile:
 
A lot easier to solve in polar form :)
 

Similar threads

Replies
17
Views
3K
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
1
Views
3K
Replies
39
Views
6K
  • · Replies 15 ·
Replies
15
Views
1K
  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 2 ·
Replies
2
Views
4K
Replies
4
Views
4K