Complex Numbers: Equation involving the argument operator.

  • Thread starter Bradyns
  • Start date
  • #1
20
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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?
 

Answers and Replies

  • #3
20
0
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
266
2
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.
 

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