complex locus - help (1 Viewer)

[fatmike93]

following on from my other question that no one could do [ integrate 1/ln(tanx) http://community.boredofstudies.org/14/mathematics-extension-2/242473/insane-maths-problem-help.html]
i have another question that you might be able to help me with

sketch the locus of arg(z-1)(z+1) = pie/3

 

[queenofdorks]

gg

 

[khorne]

Is that arg(z-1)/(z+1) = pi/3 or arg(z-1) * arg(z+1 etc.

Assuming it's the former:

arg(z-1) - arg(z+1) = pi/3

i.e arg(z-1) = pi/3 + arg(z+1)

so the vectors z+1 (from -1 to p) and z-1 (from 1 to p) are required. If we draw a parallel line to z-1, its angle must be pi/3 greater than arg(z+1). I.e this is on top of the x-axis, as if it was below, we'd be going clockwise, i.e subtracting the angle. Thus, the locus formed is a major arc of a circle, end points (not including) (-1,0) and (1,0). The angle between z+1 and z-1 is also pi/3 (that is partly the reason we know it's a major arc).

 

[Affinity]

the original question:
http://www.wolframalpha.com/input/?i=plot+sqrt(3)(x^2+-+y^2+-+1)+=+2xy
remove the parts in the 2nd and 4th qudarant

 

[fatmike93]

nah i think you misunderstodd my question

its meant to be arg(z-1)(z+) = pie/3

in other words sketch the locus of arg(z-1) + arg(z+1) = pie /3

 

[shaon0]

nah i think you misunderstodd my question

its meant to be arg(z-1)(z+) = pie/3

in other words sketch the locus of arg(z-1) + arg(z+1) = pie /3

Let z=x+iy:
arg((x-1)+iy)+arg((x+1)+iy)=pi/3
atan(y/(x-1))+atan(y/(x+1))=pi/3
tan(pi/3)=(y/(x-1)+y/(x+1))/(1-y^2/(x^2-1))
sqrt(3)=2xy/(x^2-1-y^2)
Affinity has already graphed it.

 

[fatmike93]

wow thank you

 

[queenofdorks]

nice work

i would have never thought of doing it that way!

 

[fatmike93]

shut up dorky queen

hey shano how do u do this then. basically the same thing.

arg z[z-(root3 +i)] = pie/6

 

[Affinity]

arg z[z-(root3 +i)] = pie/6

write a = sqrt(3) + i
let p+iq = w = z - a/2 = (x - sqrt(3)/2 ) + (y-1/2)i


arg( (z - a/2)^2 - a^2 /4 ) = pi/6
arg( w^2 - a^2 /4 ) = pi/6
arg( (p^2 - q^2 - 1/2) + i(2pq - sqrt(3)/2 ) = pi / 6

(2pq - sqrt(3)/2 ) / (p^2 - q^2 - 1/2) = 1/sqrt(3)

(2sqrt(3)pq - 3/2) = (p^2 - q^2 - 1/2)
q^2 + 2sqrt(3)pq - p^2 = 1
(q + sqrt(3)p)^2 - 4p^2 = 1

basically another hyperbola with some translations and then you will need to cut off part of each branch.
http://www.wolframalpha.com/input/?...+-+sqrt(3)/2+)+)^2+-+4(x+-+sqrt(3)/2+)+^2+=+1



for some reason I have this feeling that you get a kick when no one answers your questions?

 

[fatmike93]

nah, according to the phoenix text book the locus is just a line through the origin and the point (3,1) on the argand diagram and the angle is pie/6. i basically got what u got and was just checking.

stupid book.

 

[shaon0]

shut up dorky queen

hey shano how do u do this then. basically the same thing.

arg z[z-(root3 +i)] = pie/6

It's Shaon0, i've gotten something like Affinity but with x,y=/=0. It's obviously a hyperbola not a line, unless i'm making a major mistake in my working.