HSC 2015 MX2 Marathon (archive) (1 Viewer)

Kaido · Jan 1, 2015

Re: HSC 2015 4U Marathon

dan964 said:
hence show that cos 9 and sin 9 satisfy the equation
16x^5-20x^3+5x-1/sqrt(2) = 0
angles are in degrees please convert to radians

no brackets, hard to tell which polynomial is where

FrankXie · Jan 1, 2015

Re: HSC 2015 4U Marathon

Kaido said:
no brackets, hard to tell which polynomial is where

$16x^5-20x^3+5x-\frac{1}{\sqrt2}=0$

porcupinetree · Jan 1, 2015

Re: HSC 2015 4U Marathon

FrankXie said:
good try but i'm afraid it is invalid.
$|z|+|z-2i|=|z-4+2i|+|z-2i|=|4-2i-z|+|z-2i|\geq|4-2i-z+z-2i|=|4-4i|=4\sqrt2, $ where equality holds true when the vectors $ 4-2i-z $ and $ z-2i $ are in the same direction. By solving simultaneous equations ( to find the point of intersection of two lines), the minimum $ 4\sqrt2 $ occurs when $ z=\frac73-\frac13i.$

When you wrote this line:
|z|+|z-2i|=|z-4+2i|+|z-2i|
Did you deduce that |z| = |z-4+2i| because the distance from (4, -2) to the line y = 2x - 5 is equal to the distance from (0, 0) to the line? Thanks.

FrankXie · Jan 1, 2015

Re: HSC 2015 4U Marathon

porcupinetree said:
When you wrote this line:
|z|+|z-2i|=|z-4+2i|+|z-2i|
Did you deduce that |z| = |z-4+2i| because the distance from (4, -2) to the line y = 2x - 5 is equal to the distance from (0, 0) to the line? Thanks.

because that is given

Kaido · Jan 1, 2015

Re: HSC 2015 4U Marathon

9 = pi/20
we want to express in terms of 5x -> so becomes pi/4

16x^5-20x^3+5x-1/sqrt(2)
16x^5-20x^3+5x=1/sqrt(2)
LHS=cos5x/sin5x (same thing)=pi/4
=1/sqrt(2)
=RHS

(since sin(pi/4)=cos(pi/4) at pi/4)

t_davis9 · Jan 2, 2015

Re: HSC 2015 4U Marathon

NEXT QUESTION
Sorry, can't tell if the old one was finished.
http://imgur.com/k3Xgd2N
This was taken from the final page of my complex numbers test

FrankXie · Jan 2, 2015

Re: HSC 2015 4U Marathon

t_davis9 said:
NEXT QUESTION
Sorry, can't tell if the old one was finished.
http://imgur.com/k3Xgd2N
This was taken from the final page of my complex numbers test

The last part of the question is interesting. I like it.

chocolatelatte · Jan 2, 2015

Re: HSC 2015 4U Marathon

FrankXie said:
NEXT QUESTION

$$ If the complex number $ z $ satisfies that $ |z|=|z-4+2i|, $ find the minimum value of $ |z|+|z-2i|.$

this has been answered already, but i have a quick question
why can't you just find the perp distances of (0,0) and (0,2) to the locus determined using the given info, and then add them together??
thanks

InteGrand · Jan 2, 2015

Re: HSC 2015 4U Marathon

t_davis9 said:
NEXT QUESTION
Sorry, can't tell if the old one was finished.
http://imgur.com/k3Xgd2N
This was taken from the final page of my complex numbers test

$\text{Given: } z=\cos \theta +i\sin\theta$

It suffices to show all the results assuming $\theta \in \left [ -\pi,\pi \right ]$ , due to the periodicity and powers of the functions in question. So assume $\theta \in [ -\pi,\pi]$ .

(i) $1+z=(1+\cos \theta)+i\sin \theta$

$\therefore |1+z|=\sqrt{(1+\cos \theta)^2 + \sin^2 \theta}$

$=\sqrt{1^2 + 2\cos \theta + \cos^2 \theta + \sin^2 \theta}$

$=\sqrt{2+ 2\cos \theta }$ (since sin² t + cos² t = 1 ∀t ∈ ℝ)

$=\sqrt{2(1+\cos \theta )}$ .

If cos θ ≠ -1, ℝe(1+z) = 1+cos θ > 1+(-1) = 0, so that

$\arg (1+z)=\tan ^{-1}\left (\frac{\Im(1+z)}{\Re(1+z)} \right )$

$=\tan ^{-1}\left (\frac{\sin \theta}{1+\cos\theta} \right )$

$=\tan ^{-1}\left (\tan \frac{\theta}{2} \right )$ , using the identity $\frac{\sin \theta}{1+\cos \theta}=\tan \frac{\theta}{2}$ for cos θ ≠ -1 (this can be proved using t-formulae or half-angle identities)

$=\frac{\theta}{2}$ , since $\frac{\theta}{2}\in\left ( -\frac{\pi}{2},\frac{\pi}{2} \right )$ , because we assumed that $\theta \in \left [ -\pi,\pi \right ]$ , and $\cos \theta \neq -1\Rightarrow \theta \neq \pm \pi$ .

So for cos θ ≠ -1, $1+z=\sqrt{2(1+\cos\theta)}\left ( \cos\frac{\theta}{2}+i\sin\frac{\theta}{2} \right )$

If cos θ = -1 (that is, if $\theta = \pm \pi$ ), then 1+z = (1-1)+i.sin(±π) = 0, and hence arg(1+z) is undefined. However, the above formula still holds when 1+z = 0 ⇔ cos θ = -1.

So we have $1+z=\sqrt{2(1+\cos\theta)}\left ( \cos\frac{\theta}{2}+i\sin\frac{\theta}{2} \right )\text{ }\forall \theta \in [-\pi,\pi]$ .

(ii) (1+z)⁴ = 1 + 4z+6z² + 4z³ + z⁴, by the binomial theorem.

But by de Moivre's theorem and part (i),

$(1+z)^4=\left ( \sqrt{2(1+\cos\theta)}\right ) ^4\left ( \cos\frac{4\theta}{2}+i\sin\frac{4\theta}{2} \right )$

$\Rightarrow (1+z)^4=4(1+\cos \theta)^2\left ( \cos2\theta+i\sin2\theta \right )$

$=4\left (2\cos^2 \frac{\theta}{2}\right )^2 ( \cos2\theta+i\sin2\theta)$ , since $\cos \theta = 2\cos^2\frac{\theta}{2}-1\Leftrightarrow 1+\cos \theta = 2\cos^2\frac{\theta}{2}$ (from double angle identity for cosine)

$=16\cos^4 \frac{\theta}{2}\left ( \cos 2\theta+i\sin 2\theta \right )$ .

Equating both expressions for (1+z)⁴,

$1+4z+6z^2+4z^3+z^4=16\cos^4 \frac{\theta}{2}\left ( \cos 2\theta+i\sin 2\theta \right )$

The identity required to be shown follows from taking real parts of the above equation and recalling that by de Moivre's theorem, $\Re (z^k)=\cos k\theta$ and that 'real part' is a linear operator, so that the real part of a sum is the sum of real parts.

The second expression is given by $4\sin \theta + 6\sin 2\theta + 4\sin 3\theta+\sin 4\theta=16\cos^4 \left (\frac{\theta}{2}\right )\sin 2\theta \text{ }\mathbf{(Equation \text{ }(\ast))}$ , which follows in a similar way, this time by taking imaginary parts.

(iii)

The first identity follows by dividing Equation (*) by the identity required to prove in part (ii) (and recalling that $\tan X = \frac{\sin X}{\cos X}$ ), for $16\cos^4 \frac{\theta}{2}\neq 0 \Leftrightarrow \frac{\theta}{2}\neq\pm \frac{\pi}{2}\Leftrightarrow \theta \neq \pm \pi$ (so that the factor of $16\cos^4 \frac{\theta}{2}$ can be cancelled from the numerator and denominator).

For the second identity, we rearrange the identities proven in part (ii) to get:

$4\sin\theta + 4\sin 3\theta + \sin 4\theta = 16\cos^4\left ( \frac{\theta}{2}\right)\sin 2\theta - 6\sin 2\theta = \left ( 16\cos^4 \frac{\theta}{2} -6 \right )\sin 2\theta \text{ (1)}$

and

$1 + 4\cos \theta + 4\cos 3\theta + \cos4\theta = 16\cos^4\left ( \frac{\theta}{2}\right)\cos 2\theta - 6\cos2\theta = \left ( 16\cos^4 \frac{\theta}{2} -6 \right )\cos 2\theta \text{ (2)}$ .

The identity follows upon dividing (1) by (2), provided that $16\cos ^4 \frac{\theta}{2} - 6\neq 0$ .

FrankXie · Jan 3, 2015

Re: HSC 2015 4U Marathon

chocolatelatte said:
this has been answered already, but i have a quick question
why can't you just find the perp distances of (0,0) and (0,2) to the locus determined using the given info, and then add them together??
thanks

No the perpendicular distance is totally different from the distance between two points. The question actually asked us to find one point P from the line y=2x-5 so that the sum of AP and BP (where A(0,0) and B(0,2), AP and BP mean distances between points NOT perp distances) is minimized. And this is a classic geometry problem (maybe referred to as bend path probelm?), where in the solution the optimal point P is chosen such that the angle of incidence equals the angle of reflection.

FrankXie · Jan 3, 2015

Re: HSC 2015 4U Marathon

Here is my try
$1+z=1+cos\theta+i\sin\theta=1+\cos^2\frac{\theta}{2}-\sin^2\frac{\theta}{2}+2i\sin \frac{\theta}{2} \cos\frac{\theta}{2}=2\cos^2\frac{\theta}{2}+2i \sin\frac{\theta}{2}\cos\frac{\theta}{2} = 2\cos \frac{\theta}{2}{\rm cis}\frac{\theta}{2}$
$\therefore (1+z)^4=\left(2\cos\frac{\theta}{2}\right)^4\left({\rm cis}\frac{\theta}{2}\right)^4=16\cos^4\frac{\theta}{2}{\rm cis}2\theta$
$$ On the other hand, $ (1+z)^4=1+4z+6z^2+4z^3+z^4=1+4{\rm cis}\theta+6{\rm cis}2\theta+4{\rm cis}3\theta+{\rm cis}4\theta=(1+4\cos\theta+6\cos2\theta+4\cos 3\theta+\cos4\theta)+i(4\sin\theta+6\sin2\theta+4 \sin 3\theta+\sin4\theta)$
$Now equating the real part and imaginary part of these two ways of representing $ (1+z)^4, $ we arrive at$
$1+4\cos\theta+6\cos2\theta+4\cos 3\theta+\cos 4\theta=16\cos^4\frac{\theta}{2}\cos2\theta $ (1) $$
and
$4\sin\theta+6\sin2\theta+4\sin3\theta+\sin4\theta=16\cos^4\frac{\theta}{2}\sin2\theta $ (2) $$

Dividing equations (1) and (2),

$\frac{4\sin \theta+6\sin 2\theta+4\sin 3\theta+\sin4 \theta}{1+4\cos\theta+6\cos2\theta+4\cos 3\theta+\cos 4\theta}=\tan2\theta $ (3) $$

On both sides of equation (3) multiplying by the denominator,
$4\sin \theta+6\sin 2\theta+4\sin 3\theta+\sin4 \theta=\tan2\theta(1+4\cos\theta+6\cos2\theta+4\cos 3\theta+\cos 4\theta),$
$(4\sin \theta+4\sin 3\theta+\sin4 \theta)+6\sin 2\theta=\tan2\theta(1+4\cos\theta+4\cos 3\theta+\cos 4\theta)+6\tan2\theta\cos2\theta, $ from which cancelling $ 6\sin 2\theta, $ we obtain $$
$4\sin \theta+4\sin 3\theta+\sin4 \theta=\tan2\theta(1+4\cos\theta+4\cos 3\theta+\cos 4\theta)$
Therefore,
$\tan2\theta=\frac{4\sin \theta+4\sin 3\theta+\sin4 \theta}{1+4\cos\theta+4\cos 3\theta+\cos 4\theta}$

Kaido · Jan 3, 2015

Re: HSC 2015 4U Marathon

Interesting solve by Frank, esp the first part.
And looking at the amount of latex Integrand used, you can tell this guy's a beast

Immortality · Jan 3, 2015

Re: HSC 2015 4U Marathon

Find the locus of $arg(z-2) + arg(z+2) = \pi$

porcupinetree · Jan 3, 2015

Re: HSC 2015 4U Marathon

Immortality said:
Find the locus of $arg(z-2) + arg(z+2) = \pi$

The y-axis (x=0)

Kaido · Jan 3, 2015

Re: HSC 2015 4U Marathon

Rofl

FrankXie · Jan 3, 2015

Re: HSC 2015 4U Marathon

Immortality said:
Find the locus of $arg(z-2) + arg(z+2) = \pi$

$$ Let $ z=x+iy, $ then $ z-2=(x-2)+iy, z+2=(x+2)+iy. $ Taking tangent on both sides of $ {\rm arg}(z-2)+{\rm arg}(z+2)=\pi, \frac{\frac{y}{x-2}+\frac{y}{x+2}}{1-\frac{y}{x-2}\times\frac{y}{x+2}}=0, $ which is simplied to $ xy=0.$
$By inspection, the locus is the positive y-axis and the line interval $ -2\leq x<2 $ on the x-axis.$

PhysicsMaths · Jan 3, 2015

Re: HSC 2015 4U Marathon

FrankXie said:
$$ Let $ z=x+iy, $ then $ z-2=(x-2)+iy, z+2=(x+2)+iy. $ Taking tangent on both sides of $ {\rm arg}(z-2)+{\rm arg}(z+2)=\pi, \frac{\frac{y}{x-2}+\frac{y}{x+2}}{1-\frac{y}{x-2}\times\frac{y}{x+2}}=0, $ which is simplied to $ xy=0.$
$By inspection, the locus is the positive y-axis and the line interval $ -2\leq x<2 $ on the x-axis.$

Wasn't really sure what to do because i've only seen arg(z-2) - arg (z+2) = +-pi

Kaido · Jan 3, 2015

Re: HSC 2015 4U Marathon

Better just to use a geometrical representation (so much easier and is probably the preferred method anyways)

braintic · Jan 3, 2015

Re: HSC 2015 4U Marathon

FrankXie said:
$$ Let $ z=x+iy, $ then $ z-2=(x-2)+iy, z+2=(x+2)+iy. $ Taking tangent on both sides of $ {\rm arg}(z-2)+{\rm arg}(z+2)=\pi, \frac{\frac{y}{x-2}+\frac{y}{x+2}}{1-\frac{y}{x-2}\times\frac{y}{x+2}}=0, $ which is simplied to $ xy=0.$
$By inspection, the locus is the positive y-axis and the line interval $ -2\leq x<2 $ on the x-axis.$

Why not the negative y-axis?
There is no requirement that arguments must be principal values.

Just check z=-2i:

arg(z-2)+arg(z+2) = arg (-2-2i) + arg (2-2i) = 5pi/4 - pi/4 = pi

If the equation is satisfied by a complex number using any of its infinite number of arguments, then that point lies on the locus.

RealiseNothing · Jan 3, 2015

Re: HSC 2015 4U Marathon

t_davis9 said:
NEXT QUESTION
Sorry, can't tell if the old one was finished.
http://imgur.com/k3Xgd2N
This was taken from the final page of my complex numbers test

Eh a geometrical way to prove the first part (and honestly anytime you see $1+z$ , doing it geometrically is probs the way to go).

By sine rule $\frac{|1+z|}{|z|}=\frac{\sin(\theta)}{\sin(\frac{ \theta}{2})}$

But $|z|=1$ and $\sin(\theta)=2\sin(\frac{\theta}{2})(\cos(\frac{ \theta}{2})$

So $|1+z| = 2\cos(\frac{\theta}{2})$

And obviously $arg(1+z)=\frac{\theta}{2}$ so we get $1+z = 2\cos(\frac{\theta}{2})cis(\frac{\theta}{2})$

HSC 2015 MX2 Marathon (archive) (1 Viewer)

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what is that?It is Cowpea

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