Locus defined by arguments (1 Viewer)

[Porcia]

Hi guys, i got a question for y'all... help, really, moresoof. In a question as such:

arg(z-2) - arg (z+2) = pi/3
for example - how does one tell whether the locus is above or below the x-axis? As such, its converse:
arg(z+2) - arg (z-2) = pi/3;
would it most defintely be the converse of whatever side of axis the other would appear? Also, does this apply whereas:
arg(z-i) - arg (z+i) = pi/3 is used?

Thanks muchly

 

[pLuvia]

arg(z-2) - arg (z+2) = pi/3

Let's say this Let arg(z-2) be A and arg(z+2) be B hence,
arg(z-2) - arg (z+2) = pi/3
.: A - B = pi/3
.: A>B
So the locus must be above the x axis in order for this A>B to hold true

arg(z+2) - arg (z-2) = pi/3

Same here

arg(z-i) - arg (z+i) = pi/3

And here

 

[.ben]

is it always above the x axis? (except in cases of +- right?)

 

[Templar]

Start at the first point and go anticlockwise to the second, if the angle is positive.

 

[pLuvia]

Nope, for arg(z-i) - arg (z+i) = pi/3 it's on the left hand side of the y-axis
Depends on the question

is it always above the x axis? (except in cases of +- right?)