# 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?

Last edited:

#### fan96

##### 617 pages
If

$\arg(z+z^2) = \pi/3$,

then

$\tan \frac\pi 3 = \frac{\mathrm{Im}(z^2+z)}{\mathrm{Re}(z^2+z)}$

But since $\tan\pi/3 = \tan -2\pi/3$,

$\tan \frac\pi 3 = \frac{\mathrm{Im}(z^2+z)}{\mathrm{Re}(z^2+z)}$

also gives the locus of $\arg(z+z^2) =-2\pi/3$.

To get around this we can require $z^2 +z$ to be in the first quadrant, i.e. $\mathrm{Re}(z^2+z) > 0$ and $\mathrm{Im}(z^2+z) > 0$.

Setting $z = x + iy$, this gives:

$\frac{y\left(2x+1\right)}{\left(x+y\right)\left(x-y\right)+x}=\sqrt{3}, \quad \left(x+y\right)\left(x-y+1\right)+2xy\ >\ 0$

Last edited:

#### 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!