Complex Numbers: Equation involving the argument operator.

In summary, the conversation discusses solving a complex number equation using the argument and tangent functions. The simplified equation is y = (5/3)x - (1/3). The conversation also mentions a shortcut method using the gradient and a line created from two points. Both methods are considered valid for solving the equation.
  • #1
Bradyns
20
0

Homework Statement


Question:
Sr5oxE3.png


Homework Equations


Any relevant to complex numbers.

The Attempt at a Solution



Given,

Arg([itex]\frac{z}{w}[/itex])= 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([itex]\theta[/itex])=[itex]\frac{y+2}{x+1}[/itex]

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

tan([itex]\theta[/itex])=[itex]\frac{5}{3}[/itex]

RHS = LHS

[itex]\frac{y+2}{x+1}[/itex] = [itex]\frac{5}{3}[/itex]

Simplified
y = [itex]\frac{5}{3}[/itex]x -[itex]\frac{1}{3}[/itex]

Is this right?
 
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  • #2
Yes it is correct.
 
  • #3
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 = [itex]\frac{3 - (-2)}{2 - (-1)}[/itex] = [itex]\frac{5}{3}[/itex]

y = [itex]\frac{5}{3}[/itex]x + b

(-2) = [itex]\frac{5}{3}[/itex](-1) + b

-6 = -5 + 3b
-1 = 3b
b = -[itex]\frac{1}{3}[/itex]
 
Last edited:
  • #4
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.
 

1. What is a complex number?

A complex number is a number that is composed of a real part and an imaginary part. It is written in the form a + bi, where a is the real part and bi is the imaginary part (with i representing the imaginary unit).

2. What is the argument operator in complex numbers?

The argument operator in complex numbers is written as arg(z), where z is a complex number. It represents the angle between the positive real axis and the vector representing the complex number in the complex plane.

3. How do you solve equations involving the argument operator?

To solve equations involving the argument operator, you must first express the complex numbers in polar form (r(cosθ + isinθ)). You can then use trigonometric identities to manipulate the equation and solve for the variable.

4. Can the argument operator have multiple values?

Yes, the argument operator can have multiple values. This is because there are infinitely many angles that have the same tangent value, which is what the argument operator represents. When solving equations involving the argument operator, it is important to consider all possible values of the argument.

5. What is the principal value of the argument operator?

The principal value of the argument operator is the value of the argument that falls within the range -π < arg(z) ≤ π. It is considered the principal value because it is the most commonly used value when working with complex numbers and the argument operator.

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