Locus problem (complex numbers)

[willat8]

Hi! We started doing complex numbers in maths class a couple of weeks ago, and I'm not fully understanding sketching the locus of points.

Homework Statement

Sketch the locus of z:

$$arg\left(\frac{z-2}{z+2}\right) = \frac{\pi}{3}$$

The Attempt at a Solution

I've rewritten as

$$arg(z-2)-arg(z+2)$$

and have constructed a triangle with corners Re(-2), Re(2) and z.

I understand the angle subtended by the real axis must equal $$\frac{\pi}{3}$$. I do not understand the solution given by our maths teacher; that the locus of z is a kind of truncated circle above the real axis, with nothing to draw below.

 

[lanedance]

Hi! We started doing complex numbers in maths class a couple of weeks ago, and I'm not fully understanding sketching the locus of points.

Homework Statement

Sketch the locus of z:

$$arg\left(\frac{z-2}{z+2}\right) = \frac{\pi}{3}$$

The Attempt at a Solution

I've rewritten as

$$arg(z-2)-arg(z+2)$$

and have constructed a triangle with corners Re(-2), Re(2) and z.

I understand the angle subtended by the real axis must equal $$\frac{\pi}{3}$$. I do not understand the solution given by our maths teacher; that the locus of z is a kind of truncated circle above the real axis, with nothing to draw below.

so you have
$$arg(z-2)-arg(z+2) = \pi/3$$

i think the triangle you shold be drawing (above the real axis) is that defined by the points in the complex plane
z = 0 (the origin)
z-2
z+2
now as z=(a+ib),it is determined by 2 unknowns,

if you assume one, as you know the lengths of the three sides upto one unknown, (one is always 4, and the other 2 depend only z) and an angle (pi/3), then in theory you can solve for the reamaining part of z. (it could be easier to think of z in terms of a magnitude & angle)

imagine when z is purely complex, its splits the larger triangle into two right triangles and the length of z will be given by
$$tan(\pi/6)= \frac{2}{(|z|)}$$

now if you look below the real axis, the argument of (z-2) will always be less than that of (z+2) so there is no solution there

 

[willat8]

Cheers lanedance. I chatted to my maths teacher today about it, and his explanation correlated nicely to yours. I'm heading towards understanding these problems a little better.

 

[lanedance]

no worries, the best way to learn is to get in and have go as you're doing