BOS trial (1 Viewer)

mathsbrain · Aug 29, 2018

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?

Thanks in advanced!

Paradoxica · Aug 30, 2018

mrbunton said:
what's the bos trial?

ban

fan96 · Aug 30, 2018

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:

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

mathsbrain · Aug 31, 2018

Paradoxica said:
ban

what? is that banned?

Paradoxica · Sep 1, 2018

mathsbrain said:
what? is that banned?

whoosh

was a joke my dude

BLIT2014 · Sep 1, 2018

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

mathsbrain · Sep 1, 2018

Thanks for the fun!

Trebla · Sep 1, 2018

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

There are plans underway to hold them this year in early October. Stay tuned for more details.

BOS trial (1 Viewer)

mathsbrain

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fan96

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mathsbrain

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BLIT2014

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