BOS trial (1 Viewer)

mathsbrain

Member
Was stuck on a question:
Sketch the locus defined by arg(z^2+z)=pi/3, stating its equation.
Obviously this will turn into the hyperbola with the positive branch, but how do we find the equation?

Also, anyone know when the bored of studies trial will be held this year, of is it held already?

mrbunton

Member
ive never seen this questions before; but this is how i would solve it. How is it "obviously" a hyperbola btw?
argz+argz+1=pi/3
take tan of both sides, or sin or cos or whatever i dont think it matters
for tan actually: tan(a(z+1)+a(z))= sqr3
tan(arg(z))=Im/Re=y/x
tan(arg(z+1))=y/x+1
let tan(arg(z+1))=a and tan(arg z) = b

we have a+b/(1-ab)=sqr3 (angle formula)
(y/x+y/(x+1)) / (1-y^2/(x^2+x))=3^0.5
just manipulate and u should get an answer; im pretty sure this is incorrect but meh. It gives a shape that looks like at tilted hyperbola btw.

what's the bos trial?

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• BLIT2014

-insert title here-
• jathu123

fan96

617 pages
If

,

then

But since ,

also gives the locus of .

To get around this we can require to be in the first quadrant, i.e. and .

Setting , this gives:

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mathsbrain

Member
what? is that banned?

BLIT2014

The pessimistic optimist.
Moderator
BOS trial typically gets held later this year closer to HSC so it has not already been done.

mathsbrain

Member
Thanks for the fun!

Trebla

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