# Complex Numbers: Equation involving the argument operator.

Question:

## 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?

Yes it is correct.

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.