HSC 2016 MX2 Complex Numbers Marathon (archive) (1 Viewer)

math man · Apr 14, 2016

Re: HSC 2016 Complex Numbers Marathon

NEW
Prove that if a, b and c are concyclic and the circle passes through the origin then 1/a, 1/b and 1/c are collinear. Where a, b and c are complex and in quadrant one.

KingOfActing · Apr 14, 2016

Re: HSC 2016 Complex Numbers Marathon

math man said:
NEW
Prove that if a, b and c are concyclic and the circle passes through the origin then 1/a, 1/b and 1/c are collinear. Where a, b and c are complex and in quadrant one.

$Without loss of generality assume that $a, b, c$ are oriented clockwise. Since the opposite angles of a cyclic quadrilateral are supplementary: $\\\angle AOC = \angle ABC\\ \arg(\frac{a}{c}) = \arg(b-a) - \arg(b-c)\\\arg(\frac{a}{c}) + \arg(b-c) - \arg(b - a) = 0\\\arg\left(\frac{a(b-c)}{c(b-a)}\right) = 0\\\arg\left(\frac{\frac{1}{c} - \frac{1}{b}}{\frac{1}{a} - \frac{1}{b}}\right) = 0 \\\arg(\frac{1}{c}-\frac{1}{b}) - \arg(\frac{1}{a}-\frac{1}{b}) = 0\\\arg(\frac{1}{c}-\frac{1}{b}) - \arg(\frac{1}{b}-\frac{1}{a}) = \pi\\$Thus $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ fulfill the criteria for collinearity.$

wu345 · Apr 14, 2016

Re: HSC 2016 Complex Numbers Marathon

http://i.imgur.com/VOCNtil.png

InteGrand · Apr 14, 2016

Re: HSC 2016 Complex Numbers Marathon

wu345 said:
http://i.imgur.com/VOCNtil.png

If I recall correctly, this result was proved geometrically in one of the Q's in the 2015 BOS HSC 4U trial (the proof was based on the angle bisector theorem (https://en.wikipedia.org/wiki/Angle_bisector_theorem) which the paper got you to prove as an earlier part).

InteGrand · Apr 14, 2016

Re: HSC 2016 Complex Numbers Marathon

Also note that the circle they ask us to prove is the locus isn't just any old circle, it's actually the circle that has as diameter the points that internally and externally divide the line segment joining the points z₁ and z₂ in the complex plane in ratio k:l.

Paradoxica · Apr 14, 2016

Re: HSC 2016 Complex Numbers Marathon

InteGrand said:
Also note that the circle they ask us to prove is the locus isn't just any old circle, it's actually the circle that has as diameter the points that internally and externally divide the line segment joining the points z₁ and z₂ in the complex plane in ratio k:l.

Is there an elegant means of proving this is the diameter of the circle?

InteGrand · Apr 14, 2016

Re: HSC 2016 Complex Numbers Marathon

Paradoxica said:
Is there an elegant means of proving this is the diameter of the circle?

It can be proved using HSC geometry (iirc this was done in the 2015 BOS 4U Trial).

InteGrand · Apr 14, 2016

Re: HSC 2016 Complex Numbers Marathon

To save people having to go search up the paper, it can be found in the Dropbox link in this post: http://community.boredofstudies.org...trial-2015-results-documents.html#post7051009 .

Green Yoda · Apr 14, 2016

Re: HSC 2016 Complex Numbers Marathon

What value of satisfies z^2 = 7–24i ?

Paradoxica · Apr 14, 2016

Re: HSC 2016 Complex Numbers Marathon

Rathin said:
What value of satisfies z^2 = 7–24i ?

By inspection, the values are 4-3i and 3i-4

leehuan · Apr 14, 2016

Re: HSC 2016 Complex Numbers Marathon

Rathin said:
What value of satisfies z^2 = 7–24i ?

If you just started MX2 then the traditional step is just to let z=x+iy and solve

math man · Apr 16, 2016

Re: HSC 2016 Complex Numbers Marathon

NEW
If z= cis(theta) solve
z^4-2z^3+3z^2-2z+1=0

Paradoxica · Apr 16, 2016

Re: HSC 2016 Complex Numbers Marathon

math man said:
NEW
$If $z= $c$i$s$(\theta)$, solve$
$z^4-2z^3+3z^2-2z+1=0$

By inspection, the above expression is a perfect square.

$z^2-z+1=0 \Leftrightarrow \theta = 2n\pi \pm \frac{\pi}{3} \quad n \in \mathbb{Z}$

KingOfActing · Apr 16, 2016

Re: HSC 2016 Complex Numbers Marathon

math man said:
NEW
If z= cis(theta) solve
z^4-2z^3+3z^2-2z+1=0

For those who aren't Paradoxica:

$z^4 - 2z^3 + 3z^2 - 2z + 1 = 0 \\ z^2 - 2z + 3 - \frac{2}{z} + \frac{1}{z^2} = 0\\\left( z^2 + z^{-2} \right) - 2\left(z + z^{-1} \right) + 3 = 0\\ 2\cos{2\theta} - 4\cos{\theta} + 3 = 0\\ 4\cos^2{\theta} -4\cos{\theta} + 1 = 0\\\cos{\theta} = \frac{1}{2}\\\theta = 2\pi n \pm \frac{\pi}{3}, n \in \mathbb{Z}$

InteGrand · Apr 16, 2016

Re: HSC 2016 Complex Numbers Marathon

$\noindent By the way, we don't just want the solutions for $\theta$ above (in fact we don't need $\theta$ itself), we really want the solutions for $z$, which is $\cos \theta + i\sin \theta$. So we should find that $\cos \theta=\frac{1}{2}$ is a double root. The steps are reversible, so $\sin \theta = \pm \sqrt{1-\cos^2 \theta}= \pm \sqrt{1-\frac{1}{4}}= \pm \frac{\sqrt{3}}{2}$.$

$\noindent Since $z=\cos \theta + i\sin \theta$, the four roots of the given quartic equation are $\frac{1}{2}+\frac{\sqrt{3}}{2}i,\frac{1}{2}+\frac{\sqrt{3}}{2}i,\frac{1}{2}-\frac{\sqrt{3}}{2}i,\frac{1}{2}-\frac{\sqrt{3}}{2}i$ (i.e. two double roots).$

wu345 · Apr 16, 2016

Re: HSC 2016 Complex Numbers Marathon

part iii) please
http://i.imgur.com/sP2ojE1.png

KingOfActing · Apr 16, 2016

Re: HSC 2016 Complex Numbers Marathon

wu345 said:
part iii) please
http://i.imgur.com/sP2ojE1.png

$$We begin by letting $|u_1 - u_2| = u $ and $ \angle U_1OU_2 = \theta$. $\\$Label the midpoint of $U_1$ and $U_2$ as $M$. Construct the perpendicular to $U_1U_2$ through $M$. As this line is the perpendicular bisector to a chord of a circle, it passes through the centre of the circle $C$. $|MU_1| = \frac{u}{2}$, $|CU_1| = r$. $ \angle U_1CU_2 = 2\theta \Rightarrow \angle U_1CM = \theta, \angle U_1MC = 90^{\circ}\\$Hence by the right angle triangle ratio definitons of the sine function, $ \sin{\theta} = \frac{u}{2r}$. Now applying the cosine rule on triangle $U_1OU_2$ we obtain $ 2 - 2\cos\theta = u^2 \Longleftrightarrow \cos\theta = \frac{2-u^2}{2}\\$Finally, through the pythagorean identity: $\\\sin^2{\theta} + \cos^2{\theta} = 1\Longleftrightarrow\\\frac{u^2}{4r^2} + \left(\frac{2-u^2}{2}\right)^2 = 1 \Longleftrightarrow r^2 = \frac{1}{4-u^2}$
$$Let $u_1 = a + bi, u_2 = c + di$. $\\r^2 = \frac{1}{4 - (a-c)^2 - (b-d)^2} = \frac{1}{2 + 2ac + 2bd} = \frac{1}{(a+c)^2 + (b+d)^2} = \frac{1}{|u_1+u_2|^2}\\$Hence: $r = \frac{1}{|u_1+u_2|}$

parad0xica · Apr 17, 2016

Re: HSC 2016 Complex Numbers Marathon

wrong lol (my sol'n)

Drsoccerball · Apr 17, 2016

Re: HSC 2016 Complex Numbers Marathon

parad0xica said:
Warning: I forgot circle geometry

A high-school student wouldn't forget circle geometry.
Q.E.D.

parad0xica · Apr 17, 2016

Re: HSC 2016 Complex Numbers Marathon

Drsoccerball said:
A high-school student wouldn't forget circle geometry.
Q.E.D.

I have no intention of masking that part of my identity because it's already obvious. I guess your comment has convinced some people

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