HSC 2015 MX2 Marathon (archive) (8 Viewers)

FrankXie · Jan 30, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
still left mine unanswered

I knew that, just want more people get involved. here is my go:
$z^2=a^2(\cos2\theta+i\sin2\theta), z^2+a^2=a^2[(1+\cos2\theta)+i\sin2\theta]=a^2(2\cos^2\theta+2i\sin\theta\cos\theta)=2a^2 \cos \theta (cos\theta+i\sin\theta), \therefore\quad\frac{z}{z^2+a^2}=\frac{a{\rm cis}\theta}{2a^2\cos\theta{\rm cis}\theta}=\frac{1}{2a\cos\theta}$

FrankXie · Jan 30, 2015

Re: HSC 2015 4U Marathon

NEXT QUESTION

$$ Find the greatest value of $ |z| $ if the complex number $ z $ satisfies $ \left|z-\frac{4}{z}\right|=8.$

Drsoccerball · Jan 31, 2015

Re: HSC 2015 4U Marathon

FrankXie said:
I knew that, just want more people get involved. here is my go:
$z^2=a^2(\cos2\theta+i\sin2\theta), z^2+a^2=a^2[(1+\cos2\theta)+i\sin2\theta]=a^2(2\cos^2\theta+2i\sin\theta\cos\theta)=2a^2 \cos \theta (cos\theta+i\sin\theta), \therefore\quad\frac{z}{z^2+a^2}=\frac{a{\rm cis}\theta}{2a^2\cos\theta{\rm cis}\theta}=\frac{1}{2a\cos\theta}$

Very short and effective method mine was a bit shifty here it is: $z=acis\theta \\ \frac{1}{2acos\theta} \frac{z}{z} = \frac{z}{2azcos\theta} \\ = \frac{acis\theta}{2a^{2}cis\theta cos\theta} \\ = \frac{acis\theta}{(a^{2})(2cos^{2}\theta +2isin\theta cos\theta)} \\ = \frac{acis\theta}{(a^{2})(cos2\theta +1))} =RHS$

Drsoccerball · Jan 31, 2015

Re: HSC 2015 4U Marathon

FrankXie said:
NEXT QUESTION

$$ Find the greatest value of $ |z| $ if the complex number $ z $ satisfies $ \left|z-\frac{4}{z}\right|=8.$

Ill try this even though have no idea since no ones attempting..
$|z^2-4|=8|z| \\ \therefore \frac{|z^2 -4|}{8}=|z| \\ \therefore \text{max would be at 1/4?}$

Ekman · Jan 31, 2015

Re: HSC 2015 4U Marathon

FrankXie said:
NEXT QUESTION

$$ Find the greatest value of $ |z| $ if the complex number $ z $ satisfies $ \left|z-\frac{4}{z}\right|=8.$

Im going to give this a shot:

$\left|z-\frac{4}{z}\right|=8 \\* \therefore \left|z^2-4|\right = 8|z| \\* -z^2 +8z+4=0 \\* \therefore $max at z=4$ \\ $Rearrange the original equation:$\frac{|z^2-4|}{8} = |z| \\* \therefore |z| = \frac{3}{2}$

braintic · Jan 31, 2015

Re: HSC 2015 4U Marathon

FrankXie said:
NEXT QUESTION

$$ Find the greatest value of $ |z| $ if the complex number $ z $ satisfies $ \left|z-\frac{4}{z}\right|=8.$

A Geogebra sketch tells me that the locus defined by that equation is a pair of ellipses centred at the origin whose major axes are perpendicular to each other, and that the approximate answer is 8.47
I haven't done the algebra, so I don't know the exact answer.

Drsoccerball · Jan 31, 2015

Re: HSC 2015 4U Marathon

braintic said:
A Geogebra sketch tells me that the locus defined by that equation is a pair of ellipses centred at the origin whose major axes are perpendicular to each other, and that the approximate answer is 8.47
I haven't done the algebra, so I don't know the exact answer.

Well that's great didn't even learn it yet :/
$\text{The equation} x^4 - px^3 +qx^2 -rx+s=0 \text{has roots} \alpha , \beta , \gamma, \delta. \text{show that:} \\ \text{if} \alpha + \beta = \gamma +\delta \text{,then} p^3 -4pq +8r =0$

braintic · Jan 31, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
Well that's great didn't even learn it yet :/

That Geogebra sketch was just for me to get a visual on whether the answers offered so far are correct.
I still haven't attempted the question, so I can't rule out the possibility that there is another way to do the question without knowing that locus.

FrankXie · Jan 31, 2015

Re: HSC 2015 4U Marathon

FrankXie said:
NEXT QUESTION

$$ Find the greatest value of $ |z| $ if the complex number $ z $ satisfies $ \left|z-\frac{4}{z}\right|=8.$

Hint: Use the triangle inequality about complex numbers and solve an inequation to get |z|<= something.

VBN2470 · Jan 31, 2015

Re: HSC 2015 4U Marathon

FrankXie said:
I knew that, just want more people get involved. here is my go:
$z^2=a^2(\cos2\theta+i\sin2\theta), z^2+a^2=a^2[(1+\cos2\theta)+i\sin2\theta]=a^2(2\cos^2\theta+2i\sin\theta\cos\theta)=2a^2 \cos \theta (cos\theta+i\sin\theta), \therefore\quad\frac{z}{z^2+a^2}=\frac{a{\rm cis}\theta}{2a^2\cos\theta{\rm cis}\theta}=\frac{1}{2a\cos\theta}$

Alternate Solution

$$Since$\ z = a(\cos\theta + i\sin\theta) \Rightarrow |z|=a \Rightarrow |z|^2=a^2 \Rightarrow z\bar{z}=a^2. \\ \therefore \frac{z}{z^2+a^2}=\frac{z}{z^2+z\bar{z}}=\frac{1}{z+\bar{z}}=\frac{1}{2a\cos\theta}$

Ekman · Jan 31, 2015

Re: HSC 2015 4U Marathon

braintic said:
A Geogebra sketch tells me that the locus defined by that equation is a pair of ellipses centred at the origin whose major axes are perpendicular to each other, and that the approximate answer is 8.47
I haven't done the algebra, so I don't know the exact answer.

Well your approximations are very accurate:

$|z^2 - 4| = 8|z| \\ |z^2 - 4| \geq |z^2| - 4 \\ \therefore |z^2| - 4 \leq 8|z| \\ |z^2| - 8|z| - 4 \leq 0 \\ |z| \leq \frac{8+\sqrt{80}}{2} \\ |z| \leq 8.47$

Ekman · Jan 31, 2015

Re: HSC 2015 4U Marathon

VBN2470 said:
Alternate Solution

View attachment 31750

I just let out the biggest sigh ever, thank you dude

FrankXie · Jan 31, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
Well that's great didn't even learn it yet :/
$\text{The equation} x^4 - px^3 +qx^2 -rx+s=0 \text{has roots} \alpha , \beta , \gamma, \delta. \text{show that:} \\ \text{if} \alpha + \beta = \gamma +\delta \text{,then} p^3 -4pq +8r =0$

$\sum\alpha=\alpha+\beta+\gamma+\delta=p\Rightarrow\alpha+\beta=\gamma+\delta=\frac{p}{2}$

$\sum\alpha\beta=(\alpha+\beta)(\gamma+\delta) + \alpha \beta+\gamma \delta=q\Rightarrow\frac{p}{2}\cdot\frac{p}{2} + \alpha \beta + \gamma\delta=q\Rightarrow\alpha\beta + \gamma\delta=q-\frac{p^2}{4}$

$\sum\alpha\beta\gamma=\alpha\beta(\gamma+\delta)+(\alpha+\beta)\gamma\delta=r\Rightarrow\frac{p}{2} \alpha \beta+\frac{p}{2} \gamma \delta=r\Rightarrow\frac{p}{2}(q-\frac{p^2}{4})=r\Rightarrow p^3-4pq+8r=0$

braintic · Feb 1, 2015

Re: HSC 2015 4U Marathon

Ekman said:
Well your approximations are very accurate:

$|z^2 - 4| = 8|z| \\ |z^2 - 4| \geq |z^2| - 4 \\ \therefore |z^2| - 4 \leq 8|z| \\ |z^2| - 8|z| - 4 \leq 0 \\ |z| \leq \frac{8+\sqrt{80}}{2} \\ |z| \leq 8.47$

That actually makes a lot of sense in terms of my diagram.

Your triangle inequality in line 2 will turn into an equality when z lies on the real axis (outside -2 and 2), and that is where the points of maximum modulus were located.

Drsoccerball · Feb 1, 2015

Re: HSC 2015 4U Marathon

New question

$\text{Use the binomial expansion of}(cos\theta +isin\theta)^4 \text{With de Moivre's theorem to solve:}\\ 16x^4 -16x^2 +3=0$

Drsoccerball · Feb 1, 2015

Re: HSC 2015 4U Marathon

Can we please not let this thread die

Kaido · Feb 1, 2015

Re: HSC 2015 4U Marathon

calm down boy, this thread wont die as long as im here

Kaido · Feb 1, 2015

Re: HSC 2015 4U Marathon

cos4θ=Re ∑k=0->4 (4Ck)(cosθ)^k(isinθ)^(4−k)=8cos^4θ-8cos^2θ+1
let cosθ=x and then cos4θ=1/2
so, 8x^4-8x^2+1=1/2
16x^4-16x^2+1=0

solve cos4θ=1/2
therefore: +/- cospi/12, +/- cos5pi/12

Kaido · Feb 1, 2015

Re: HSC 2015 4U Marathon

In the meantime, someone try solving this (just curious):
x^5+x=1, find real x

Drsoccerball · Feb 1, 2015

Re: HSC 2015 4U Marathon

Kaido said:
cos4θ=Re ∑k=0->4 (4Ck)(cosθ)^k(isinθ)^(4−k)=8cos^4θ-8cos^2θ+1
let cosθ=x and then cos4θ=1/2
so, 8x^4-8x^2+1=1/2
16x^4-16x^2+1=0

solve cos4θ=1/2
therefore: +/- cospi/12, +/- cos5pi/12

Makes sense the question says 3 though so in that case you'd make $cos4\theta =\frac{-1}{2}$ Thanks

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