HSC 2015 MX2 Marathon (archive) (1 Viewer)

VBN2470 · Feb 4, 2015

Re: HSC 2015 4U Marathon

FrankXie said:
NEXT QUESTION

$You may use $ \cos3\theta=4\cos^3\theta-3\cos\theta. $ Solve the equation $ 4x^3-9x-1=0. $ Leave your answer in exact trigonometric form.$

Don't have time to completely do it now, but set $$x=\sqrt{3}\cos\theta$$ and use the given identity to find the solutions to the equation.

FrankXie · Feb 4, 2015

Re: HSC 2015 4U Marathon

NEXT QUESTION

$(i) Factorise $ z^8+1 $ into real quadratic factors.$
$$ (ii) Hence show that $ \cos4\theta=8\left(\cos^2\theta-\cos^2\frac{\pi}{8}\right)\left(\cos^2\theta-\cos^2\frac{3\pi}{8}\right).$

VBN2470 · Feb 4, 2015

Re: HSC 2015 4U Marathon

NEW QUESTION:

$$ Find the Cartesian form of the complex number whose modulus and argument is $3$ and $\frac{\pi}{8}$ respectively.$

Drsoccerball · Feb 4, 2015

Re: HSC 2015 4U Marathon

VBN2470 said:
NEW QUESTION:

$$ Find the Cartesian form of the complex number whose modulus and argument is $3$ and $\frac{\pi}{8}$ respectively.$

$$ Note that$ \frac{(cos2\theta +1)}{2}=cos^2\theta \\ $ real part $ = 3cos\frac{\pi}{8} \\$Imaginary part$ =3sin\frac{\pi}{8}\\ $Equate the equation of* cos\theta \\ cos\theta = \sqrt{\frac{1+\sqrt{2}}{2}} \\ \therefore = 3\sqrt{\frac{1+ \sqrt{2}}{2}} +3i\sqrt{\frac{1+\sqrt{2}}{2}}$

Ekman · Feb 4, 2015

Re: HSC 2015 4U Marathon

VBN2470 said:
NEW QUESTION:

$$ Find the Cartesian form of the complex number whose modulus and argument is $3$ and $\frac{\pi}{8}$ respectively.$

The working out is a pain for me to write out in LaTeX, but its basically using prelim extension 1 trig knowledge. I got:

$\frac{3}{2}\sqrt{\sqrt{2} +2} + i\frac{3}{2}\sqrt{\frac{2}{\sqrt{2}+2}}$

VBN2470 · Feb 4, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
$$ Note that$ \frac{(cos2\theta +1)}{2}=cos^2\theta \\ $ real part $ = 3cos\frac{\pi}{8} \\$Imaginary part$ =3sin\frac{\pi}{8}\\ $Equate the equation of* cos\theta \\ cos\theta = \sqrt{\frac{1+\sqrt{2}}{2}} \\ \therefore = 3\sqrt{\frac{1+ \sqrt{2}}{2}} +3i\sqrt{\frac{1+\sqrt{2}}{2}}$

Yep, pretty much the right idea, but I think you mixed up some plus signs with minus signs. Answer in simplest form is $\! $ $z=\frac{3}{2}\left[\sqrt{2+\sqrt2}+i\sqrt{2-\sqrt2} \right]$ $$ .

VBN2470 · Feb 4, 2015

Re: HSC 2015 4U Marathon

NEW QUESTION:

$$ Show that if for any complex number $z$ with $|z|=1$, then Re\left(\frac{1-z}{1+z} \right)=0$

Ekman · Feb 4, 2015

Re: HSC 2015 4U Marathon

VBN2470 said:
NEW QUESTION:

$$ Show that if for any complex number $z$ with $|z|=1$, then Re\left(\frac{1-z}{1+z} \right)=0$

$$We know that $z \times \bar{z} = 1 $ (given in the question)$

$\frac{1-z}{1+z} \times \frac{1+\bar{z}}{1+\bar{z}}$

$=\frac{\bar{z}-z}{2+\bar{z}+z}$

$\bar{z} - z $ is purely imaginary $$

$\therefore \frac{\bar{z}-z}{2+\bar{z}+z} $ is purely imaginary$$

$\therefore $ since $ \frac{\bar{z}-z}{2+\bar{z}+z} $ is purely imaginary then $Re(\frac{1-z}{1+z})=0$

Drsoccerball · Feb 4, 2015

Re: HSC 2015 4U Marathon

VBN2470 said:
NEW QUESTION:

$$ Show that if for any complex number $z$ with $|z|=1$, then Re\left(\frac{1-z}{1+z} \right)=0$

$Let $z= x+iy $sub it in the equation$ $Multiply the denominater by its conjugate and the top by the denominators conjugate$ \\ $After equation real parts you get:$ \frac{1-(x^2+y^2)}{1+2x +x^2 +y^2} $numerator$= 1-1=0 \\ \therefore=0 $Sorry for the lack of working out and the silly mistakes in the other question im just really tired$

braintic · Feb 5, 2015

Re: HSC 2015 4U Marathon

VBN2470 said:
NEW QUESTION:

$$ Show that if for any complex number $z$ with $|z|=1$, then Re\left(\frac{1-z}{1+z} \right)=0$

The geometrical solution is much easier.
The vectors representing 1-z and 1+z are perpendicular (diagonals of a rhombus), meaning their quotient has an argument of (+/-)pi/2, meaning the quotient is purely imaginary.

Ekman · Feb 5, 2015

Re: HSC 2015 4U Marathon

FrankXie said:
NEXT QUESTION

$(i) Factorise $ z^8+1 $ into real quadratic factors.$
$$ (ii) Hence show that $ \cos4\theta=8\left(\cos^2\theta-\cos^2\frac{\pi}{8}\right)\left(\cos^2\theta-\cos^2\frac{3\pi}{8}\right).$

Can someone solve this please, I got something weird for ii. I got:

$\cos^2(4\theta) = 8cos^4(\theta)\left(\cos^2(\theta)-\cos^2(\frac{\pi}{8})\right)\left(\cos^2(\theta) - \cos^2(\frac{3\pi}{8})\right)$

HeroicPandas · Feb 5, 2015

Re: HSC 2015 4U Marathon

Ekman said:
Can someone solve this please, I got something weird for ii. I got:

$\cos^2(4\theta) = 8cos^4(\theta)\left(\cos^2(\theta)-\cos^2(\frac{\pi}{8})\right)\left(\cos^2(\theta) - \cos^2(\frac{3\pi}{8})\right)$

$z^8 + 1 = \left(z^2 - 2zcos\left ( \frac{\pi}{8} \right ) + 1 \right) \left(z^2 - 2zcos\left ( \frac{3\pi}{8} \right ) + 1 \right) \left(z^2 - 2zcos\left ( \frac{5\pi}{8} \right ) + 1 \right) \left(z^2 - 2zcos\left ( \frac{7\pi}{8} \right ) + 1 \right)$

Hello, what did you do after this? (to find out where you made the mistake)

Spoiler...
ANSWER TO (ii): Check LHS, we want z^4 + z^(-4) so divide both sides by z^4

Ekman · Feb 5, 2015

Re: HSC 2015 4U Marathon

HeroicPandas said:
$z^8 + 1 = \left(z^2 - 2zcos\left ( \frac{\pi}{8} \right ) + 1 \right) \left(z^2 - 2zcos\left ( \frac{3\pi}{8} \right ) + 1 \right) \left(z^2 - 2zcos\left ( \frac{5\pi}{8} \right ) + 1 \right) \left(z^2 - 2zcos\left ( \frac{7\pi}{8} \right ) + 1 \right)$

Hello, what did you do after this? (to find out where you made the mistake)

Spoiler...
ANSWER TO (ii): Check LHS, we want z^4 + z^(-4) so divide both sides by z^4

Actually for (i) I got:
$z^8+1 = (z^2-2z\cos\frac{\pi}{8} + 1)(z^2+2z\cos\frac{\pi}{8} + 1)(z^2-2z\cos\frac{3\pi}{8} +1)(z^2+2z\cos\frac{3\pi}{8}+1)$

$And so for (ii), i subbed in $z=cis\theta$ and then i used De Morivre's theorem, equated all real parts and then double angle formula. Afterwards i expanded to get what i previously posted before.$

Zlatman · Feb 5, 2015

Re: HSC 2015 4U Marathon

I managed to get an answer to it, but there are probably better ways to do it
http://imgur.com/CqsLJB6
(still new to this, haha)

Ekman · Feb 5, 2015

Re: HSC 2015 4U Marathon

Hang on...

$\frac{z^8+1}{z^4} = z^4+\frac{1}{z^4} = 2\cos(4\theta)$

$z^8+1 = 16cos^4(\theta)\left(\cos^2(\theta)-\cos^2(\frac{\pi}{8})\right)\left(\cos^2(\theta) - \cos^2(\frac{3\pi}{8})\right)$

$2cos(4\theta) = 16\left(\cos^2(\theta)-\cos^2(\frac{\pi}{8})\right)\left(\cos^2(\theta) - \cos^2(\frac{3\pi}{8})\right)$

$cos(4\theta) = 8\left(\cos^2(\theta)-\cos^2(\frac{\pi}{8})\right)\left(\cos^2(\theta) - \cos^2(\frac{3\pi}{8})\right)$

EDIT:
$I do admit, i did skip a few steps, but for those who are still wondering how i got rid of $ i $ when i subbed in $z=cis\theta $, i only just equated all real parts$

HeroicPandas · Feb 5, 2015

Re: HSC 2015 4U Marathon

Ekman said:
Actually for (i) I got:
$z^8+1 = (z^2-2z\cos\frac{\pi}{8} + 1)(z^2+2z\cos\frac{\pi}{8} + 1)(z^2-2z\cos\frac{3\pi}{8} +1)(z^2+2z\cos\frac{3\pi}{8}+1)$

yeh got that too after using cosine supplementary angles

Drsoccerball · Feb 5, 2015

Re: HSC 2015 4U Marathon

NEW QUESTION:
$$In the expansion of$ (a+b+c+d)^{12} $What is the coefficient of the term in,$ ac^5d^6$

braintic · Feb 5, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
NEW QUESTION:
$$In the expansion of$ (a+b+c+d)^{12} $What is the coefficient of the term in,$ ac^5d^6$

It would be 12!/(5!6!).
It is the same as counting the 12 letter words formed from one A, 5 Cs and 6 Ds.

Drsoccerball · Feb 5, 2015

Re: HSC 2015 4U Marathon

Next question guys:
$$Find the product of the five 5th roots of$ \ (1+\sqrt{2}i)^3$

VBN2470 · Feb 5, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
Next question guys:
$$Find the product of the five 5th roots of$ \ (1+\sqrt{2}i)^3$

$$ If we want to solve the polynomial equation $z^{5}-\(1+\sqrt{2}i)^3=0$, then our roots will satisfy the equation $(z-\alpha)(z-\beta)(z-\gamma)(z-\delta)(z-\epsilon)=0$, where $\alpha, \beta, \gamma, \delta, \epsilon$ are the fifth roots of $\(1+\sqrt{2}i)^3$. From 3U Polynomials, the product of roots for a quintic equation $az^{5}+bz^{4}+cz^{3}+dz^{2}+ez+f=0$ will be given as $-\frac{f}{a}$, so our product will be $\(1+\sqrt{2}i)^3$. \\ NOTE that in this case $a=1$ and that $b,c,d,e$ and $f$ are not all real.$

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