Complex Numbers: Equation involving the argument operator.

Click For Summary
SUMMARY

The discussion focuses on solving a complex number equation involving the argument operator, specifically Arg(z/w) = Arg(z) - Arg(w). The participants analyze the equation using two complex numbers, z1 = -1 - 2i and z2 = 2 + 3i, leading to the conclusion that the relationship between the arguments simplifies to y = (5/3)x - (1/3). A shortcut method is also introduced, utilizing the gradient between the points z1 and z2, confirming the correctness of the original solution while suggesting that practice will enhance efficiency in both methods.

PREREQUISITES
  • Understanding of complex numbers and their properties
  • Familiarity with the argument operator in complex analysis
  • Knowledge of tangent functions and their application in geometry
  • Basic skills in algebraic manipulation and solving equations
NEXT STEPS
  • Study the properties of complex numbers in polar form
  • Learn about the geometric interpretation of complex number operations
  • Explore advanced techniques for solving complex equations
  • Investigate the use of gradients in coordinate geometry
USEFUL FOR

Students studying complex analysis, mathematicians interested in geometric interpretations, and educators teaching complex number operations.

Bradyns
Messages
20
Reaction score
0

Homework Statement


Question:
Sr5oxE3.png


Homework Equations


Any relevant to complex numbers.

The Attempt at a Solution



Given,

Arg(\frac{z}{w})= Arg(z)-Arg(w)

z=x+yi
z1 = -1-2i
z2 = 2+3i

Arg(z-z1)=Arg(z2-z1)

LHS:
Arg(x+yi+1+2i)
Arg((x+1) + i(y+2))

tan(\theta)=\frac{y+2}{x+1}

RHS
Arg(2+3i+1+2i)
Arg(3+5i)

tan(\theta)=\frac{5}{3}

RHS = LHS

\frac{y+2}{x+1} = \frac{5}{3}

Simplified
y = \frac{5}{3}x -\frac{1}{3}

Is this right?
 
Physics news on Phys.org
Yes it is correct.
 
Sunil Simha said:
Yes it is correct.

Cool beans,

Is there a simple way to solve these though?

A shortcut method?Wait, I found it...

It's the line created from the points z1= -1-2i and z2 = 2+3i

Gradient = \frac{3 - (-2)}{2 - (-1)} = \frac{5}{3}

y = \frac{5}{3}x + b

(-2) = \frac{5}{3}(-1) + b

-6 = -5 + 3b
-1 = 3b
b = -\frac{1}{3}
 
Last edited:
Yes. That's how I solved it. Your original method is still fine though. If you have loads of practice,, you'll be able to use both the methods equally efficiently.
 

Similar threads

  • · Replies 19 ·
Replies
19
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 14 ·
Replies
14
Views
2K
Replies
6
Views
3K
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 20 ·
Replies
20
Views
2K
Replies
39
Views
6K
  • · Replies 6 ·
Replies
6
Views
2K