What is the Locus of z When arg(z-1) and arg(z+1) are Both Equal to 3pi/4?

[rock.freak667]

Homework Statement

If arg(z-1)=arg(z+1)=\frac{3 \pi}{4}, find the locus of z.

Homework Equations

The Attempt at a Solution

arg(z-1)=arg(z-[1+0i]) => z lies on half the line through the point (1,0) excluding (1,0), inclined at alpha

Similarly arg(z+1) =>z lies on half the line through (-1,0),excluding (-1,0).Inclined at beta.

If I draw those two on the same graph, they form a triangle. But how do I incorporate the 3pi/4 ?

or was I supposed to draw arg(z-1)=3pi/4?

 

[tiny-tim]

Hi rock.freak667!

If arg(z-1)=arg(z+1)=\frac{3 \pi}{4}, find the locus of z.

That doesn't look right to me.

Are you sure it isn't arg(z-1) - arg(z+1) = 3π/4 ?

 

[rock.freak667]

Well it actually could be that because it's in someone's handwriting.

But if it was arg(z-1) - arg(z+1) = 3π/4.

How would I go about it?


EDIT:

I can sketch

arg(z-z_0)= \lambda


where z_0 is a fixed complex number and lambda is the argument.

 

[tiny-tim]

Well it actually could be that because it's in someone's handwriting.

But if it was arg(z-1) - arg(z+1) = 3π/4.

How would I go about it?

dot-product or just trigonometry?

 

[rock.freak667]

ahh nevermind I figured it out.

And the correct question is:

arg(z-1)-arg(z+1)=\frac{\pi}{4}