# 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?

#### fan96

##### 617 pages
If

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

then

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

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

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

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

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

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

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

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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!