Complex Numbers locus :O (1 Viewer)

[youngminii]

Okay so my teacher told us how to do this
Except she's a crap teacher
So help please
z is a set of points that moves so that Arg|z - 2| = Arg|z + 3i|
Thanks
Also, when you draw the actual locus for something like Arg|z| = pi/3
Do you leave the starting point (in this case, origin) as an open circle? Since the angle between the origin and the positive x axis is.. Well.. It doesn't exist?

 

[-tal-]

Okay so my teacher told us how to do this
Except she's a crap teacher
So help please
z is a set of points that moves so that Arg|z - 2| = Arg|z + 3i|
Thanks
Also, when you draw the actual locus for something like Arg|z| = pi/3
Do you leave the starting point (in this case, origin) as an open circle? Since the angle between the origin and the positive x axis is.. Well.. It doesn't exist?

What's the question?

And yeah, technically more correct if you do the open circle

 

[youngminii]

The question is.. Sketch and describe it
(z = x + iy obviously)

 

[-tal-]

The question is.. Sketch and describe it
(z = x + iy obviously)

meh, I'm old and out of touch.

Anyhow:
Arg|z - 2| = Arg|z + 3i|
Arg|z - 2| - Arg|z + 3i| = 0
Arg|(z - 2)/(z + 3i)| = 0

Which looks like ... see attachment

 

[youngminii]

I love you
Open circles though, right?

 

[-tal-]

Yeah, open circles.

 

[youngminii]

Can you explain exactly why? Like, geometrically (I don't understand the algebra either, like how Arg|(z - 2)/(z + 3i)| = 0 turns into the graph..)
Thanks

If it's not too much of a bother

 

[-tal-]

Arg|z - 2| = Arg|z + 3i|
Arg|z - 2| - Arg|z + 3i| = 0
Arg|(z - 2)/(z + 3i)| = 0

See those blue numbers? Think of them like a quadratic eqn eg (x - 2)(x + 3) = 0. So from Arg|(z - 2)/(z + 3i)| = 0, you get 2 points, 2 & 3i. Shove on graph.

And since the angle is 0, it's a straight line, because there's no angle for the locus of z from those points. Which, if there was an angle eg pi/3 it'd be more circular, and as it approaches 0, it'll become flatter - ie more like a line.

It's dotted because if it wasn't, the angle between 2 and -3i would be pi, which it's not supposed to be.

 

[tommykins]

Open circles is due to technically a 'point' cannot haev an argument.

 

[youngminii]

Arg|z - 2| = Arg|z + 3i|
Arg|z - 2| - Arg|z + 3i| = 0
Arg|(z - 2)/(z + 3i)| = 0

See those blue numbers? Think of them like a quadratic eqn eg (x - 2)(x + 3) = 0. So from Arg|(z - 2)/(z + 3i)| = 0, you get 2 points, 2 & 3i. Shove on graph.

And since the angle is 0, it's a straight line, because there's no angle for the locus of z from those points. Which, if there was an angle eg pi/3 it'd be more circular, and as it approaches 0, it'll become flatter - ie more like a line.

It's dotted because if it wasn't, the angle between 2 and -3i would be pi, which it's not supposed to be.

Jebus

Okay then, just for clarification, can you sketch Arg|(z - 2)/(z + 3i)| = pi/3

 

[tommykins]

Yes. its abit smaller than a semi circle from (0,-3) to (2,0) anti clockwise. (subtends pi/3 rad)

 

[-tal-]

Jebus

Okay then, just for clarification, can you sketch Arg|(z - 2)/(z + 3i)| = pi/3

here you go

 

[youngminii]

I love you.
Wait didn't tommykins say it was smaller than a semi circle? ie. minor arc? O.O -MINDFUCKED-

 

[-tal-]

I love you.
Wait didn't tommykins say it was smaller than a semi circle? ie. minor arc? O.O -MINDFUCKED-

yeah yeah, it's meant to be. paint sucks and so does complex numbers so... live with it.

 

[Continuum]

I love you.
Wait didn't tommykins say it was smaller than a semi circle? ie. minor arc? O.O -MINDFUCKED-

It doesn't matter, as long as you show the blank dots at 2 and -3i, as well as it going anticlockwise.

 

[youngminii]

Oh I see
Thank you all very, very much