MX2 Marathon (1 Viewer)

InteGrand · Jan 10, 2018

Re: HSC 2018 MX2 Marathon

altSwift said:
Find an expression for cos^4x in terms of cos4x and cos2x

Edit: The question is from terry lee so I'm guessing it implies that you use 4U complex techniques rather than 3U, I'm not sure if you can even use 3U for this...

$\noindent \textbf{Hints.}$

$\noindent $\bullet$ A 3U method: you already know a formula for cos-squared:$

$\cos^{2}\theta = \frac{1}{2}\left(1 + \cos 2 \theta\right).$

$\noindent (This is known as a \textsl{half-angle formula}.) So to get $\cos^{4}x$, try squaring both sides of the above. You will need to use the half-angle formula once more to get to the final answer.$

$\noindent $\bullet$ A 4U method: if we write $z = \cos x + i \sin x$, then we have (make sure you can prove these)$

$\cos x = \frac{z + z^{-1}}{2} \quad \text{and} \quad z^{n} + z^{-n} = 2\cos nx \quad \text{for all integers }n. \quad (\star)$

$\noindent Therefore,$

$\begin{align*}\cos^{4} x &= \left(\cos x\right)^{4} \\ &= \left(\frac{z + z^{-1}}{2}\right)^{4}\\ &= \ldots .\end{align*}$

$\noindent See if you can complete the rest, with the help of the formulas in $(\star)$.$

Drongoski · Jan 10, 2018

Re: HSC 2018 MX2 Marathon

Therefore, using 3U method:

$4cos^4 x =(2cos^2 x)^2 = (1+cos 2x)^2 = 1 + 2cos 2x + cos^2 2x \\ \\ = 1 + 2cos 2x + \frac {1}{2} (1 + cos 4x) = 1.5 + 2cos 2x + 0.5 cos 4x \\ \\ \therefore cos^4 x = \cdots$

dan964 · Jan 11, 2018

Re: HSC 2018 MX2 Marathon

CapitalSwine said:
What is the modulus and argument of z in this case? I got an answer that was different to the one in the textbook.

Sent from my Redmi Note 4 using Tapatalk

It is is circle of radius $2\sqrt{2}$ centred at 4+4i.

The construction is as follows
Join the points 4+4i (centre C) and origin O, extend this line so that it touches the circle again at B. This line OB will give us the range of values for |z|

To find the arg z, construct a diagram, observe the points E and F where the arg z is max and min, form a right angled triangle. The ratios of the sides will give the external angles, and from there using symmetry, the angles formed by the tangent at E and F at the origin, gives the max and min of the arg z.

In this problem |z| is from $2\sqrt{2} \leq |z| \leq 6\sqrt{2}$
and arg z is from $\frac{\pi}{12} \leq \arg z \leq \frac{5\pi}{12}$

altSwift · Jan 20, 2018

Re: HSC 2018 MX2 Marathon

InteGrand said:
$\noindent \textbf{Hints.}$

$\noindent $\bullet$ A 3U method: you already know a formula for cos-squared:$

$\cos^{2}\theta = \frac{1}{2}\left(1 + \cos 2 \theta\right).$

$\noindent (This is known as a \textsl{half-angle formula}.) So to get $\cos^{4}x$, try squaring both sides of the above. You will need to use the half-angle formula once more to get to the final answer.$

$\noindent $\bullet$ A 4U method: if we write $z = \cos x + i \sin x$, then we have (make sure you can prove these)$

$\cos x = \frac{z + z^{-1}}{2} \quad \text{and} \quad z^{n} + z^{-n} = 2\cos nx \quad \text{for all integers }n. \quad (\star)$

$\noindent Therefore,$

$\begin{align*}\cos^{4} x &= \left(\cos x\right)^{4} \\ &= \left(\frac{z + z^{-1}}{2}\right)^{4}\\ &= \ldots .\end{align*}$

$\noindent See if you can complete the rest, with the help of the formulas in $(\star)$.$

I understand the 3U method, and I understand the proof for the 4U method, I'm just stumped as to how ur supposed to continue with the 4U method, I seem to be going around in circles lmao

fluffchuck · Jan 20, 2018

Re: HSC 2018 MX2 Marathon

altSwift said:
I understand the 3U method, and I understand the proof for the 4U method, I'm just stumped as to how ur supposed to continue with the 4U method, I seem to be going around in circles lmao

$\noindent After expanding the expression given by InteGrand, you should get an expression in terms of $z^2$, $z^{-2}$, $z^4$ and $z^{-4}$. From here, you should try to include ($\star$) somehow. To do this, you would need some terms in the form $z^n+z^{-n}$, such that $n$ is an integer. From what we have, we can pair up $z^2$ and $z^{-2}$ (this is $n=2$), then pair up $z^4$ and $z^{-4}$ (this is $n=4$) etc.$

dan964 · Jan 21, 2018

Re: HSC 2018 MX2 Marathon

next question:

using demoivre's theorem or some other complex number theorems, find the exact value of cos 36 degrees.
go...

altSwift · Jan 21, 2018

Re: HSC 2018 MX2 Marathon

fluffchuck said:
$\noindent After expanding the expression given by InteGrand, you should get an expression in terms of $z^2$, $z^{-2}$, $z^4$ and $z^{-4}$. From here, you should try to include ($\star$) somehow. To do this, you would need some terms in the form $z^n+z^{-n}$, such that $n$ is an integer. From what we have, we can pair up $z^2$ and $z^{-2}$ (this is $n=2$), then pair up $z^4$ and $z^{-4}$ (this is $n=4$) etc.$

ah that makes more sense. we haven't done binomial expansions yet so looks like ill have to do it the long way haha

mrbunton · Feb 4, 2018

Re: HSC 2018 MX2 Marathon

z= x+iy, w = u+iv; w = z -1/z. , find locus of w if |z|=2
edit:from patel textbook

mrbunton · Feb 4, 2018

Re: HSC 2018 MX2 Marathon

si2136 said:
https://i.imgur.com/DOI8Tanr.jpg

Hopefully im right, correct me if wrong

u didnt fully complete it; the hardest part is finding the relationship between v and w.

fan96 · Feb 4, 2018

Re: HSC 2018 MX2 Marathon

mrbunton said:
u didnt fully complete it; the hardest part is finding the relationship between v and w.

$w = \frac{3x}{4} + \frac{5y}{4}i$

$\begin{cases} u = \frac{3x}{4} \\v = \frac{5y}{4}\end{cases}$

Squaring, we have

$\begin{cases} u^2 = \frac{9}{16}x^2 \\v^2 = \frac{25}{16}y^2\end{cases}$

$x^2 = \frac{16}{9}u^2, \, \text{and}\, y^2 = \frac{16}{25}v^2$

$|z| = 2 \implies x^2 + y^2 = 4$

$\therefore \frac{16}{9}u^2 + \frac{16}{25}v^2 = 4$

Replacing variables,

$\frac{16}{9}x^2 + \frac{16}{25}y^2 = 4$

$\frac{x^2}{\left(\frac{3}{2}\right)^2} + \frac{y^2}{\left(\frac{5}{2}\right)^2} = 1$ or $100x^2 + 36y^2 = 225$

Is that right?

mrbunton · Feb 4, 2018

Re: HSC 2018 MX2 Marathon

yup

mrbunton · Feb 6, 2018

Re: HSC 2018 MX2 Marathon

polynomial/complex number question:
cot(x)= y+1
a,b are roots of z^2-2z+2=0
prove that ( (y+a)^n - (y+b)^n ) / (a-b) = sin(nx) / sinx^n

fan96 · Feb 9, 2018

Re: HSC 2018 MX2 Marathon

This is an interesting question I found in the SGS notes.

If $z$ is a complex number, find the locus of $z$ if

$\mathrm{Re}\left(z-\frac{1}{z}\right) = 0$

Paradoxica · Feb 9, 2018

Re: HSC 2018 MX2 Marathon

fan96 said:
This is an interesting question I found in the SGS notes.

If $z$ is a complex number, find the locus of $z$ if

$\mathrm{Re}\left(z-\frac{1}{z}\right) = 0$

Polar bash (mod-arg for the high schoolers) makes this problem trivial...

mrbunton · Feb 9, 2018

Re: HSC 2018 MX2 Marathon

fan96 said:
This is an interesting question I found in the SGS notes.

If $z$ is a complex number, find the locus of $z$ if

$\mathrm{Re}\left(z-\frac{1}{z}\right) = 0$

=Re(x+iy -(x-iy)/x^2+y^2)=0
x - x/(x^2+y^2) = 0
times by x^2 + y^2 then divide by x;
0= x^2+y^2 -1;
1 = x^2 + y^2

fan96 · Feb 10, 2018

Re: HSC 2018 MX2 Marathon

mrbunton said:
=Re(x+iy -(x-iy)/x^2+y^2)=0
x - x/(x^2+y^2) = 0
times by x^2 + y^2 then divide by x;
0= x^2+y^2 -1;
1 = x^2 + y^2

$x^2 + y^2 = 1$ is not the correct answer.

(Be careful of any restrictions you place on the locus)

altSwift · Feb 10, 2018

Re: HSC 2018 MX2 Marathon

fan96 said:
This is an interesting question I found in the SGS notes.

If $z$ is a complex number, find the locus of $z$ if

$\mathrm{Re}\left(z-\frac{1}{z}\right) = 0$

$Noting that z \neq 0$

$z - \frac{1}{z} = z - \frac{\bar z}{z\bar z}$

$= x + iy - \frac{x - iy}{x^2 + y^2}$

$= \frac{x(x^2 + y^2 )+ iy(x^2 + y^2) -x + iy}{x^2 + y^2}$

$= \frac{x(x^2 + y^2 -1) + iy(x^2 + y^2 + 1)}{x^2 + y^2}$

$If $\mathrm{Re}\left(z-\frac{1}{z}\right) = 0$,$

$\frac{x(x^2 + y^2 -1)}{x^2 + y^2} = 0$

$x^2 + y^2 = 1$

$\therefore $ The locus of z is the unit circle excluding (0, 0)$$

altSwift · Feb 10, 2018

Re: HSC 2018 MX2 Marathon

$From a ruse paper:$

$If $z = \cos\theta + i\sin\theta$, prove that $\frac{2}{1 + z} = 1 - i\tan\left(\frac{\theta}{2}\right)$$

fan96 · Feb 10, 2018

Re: HSC 2018 MX2 Marathon

altSwift said:
$Noting that z \neq 0$

$z - \frac{1}{z} = z - \frac{\bar z}{z\bar z}$

$= x + iy - \frac{x - iy}{x^2 + y^2}$

$= \frac{x(x^2 + y^2 )+ iy(x^2 + y^2) -x + iy}{x^2 + y^2}$

$= \frac{x(x^2 + y^2 -1) + iy(x^2 + y^2 + 1)}{x^2 + y^2}$

$If $\mathrm{Re}\left(z-\frac{1}{z}\right) = 0$,$

$\frac{x(x^2 + y^2 -1)}{x^2 + y^2} = 0$

$x^2 + y^2 = 1$

$\therefore $ The locus of z is the unit circle excluding (0, 0)$$

As I said above, that's not the correct answer.

The locus is not $x^2 + y^2 = 1$ .

There are certain values of $z$ (specifically $x$ ) that need to be considered.

(By the way, $(0,0)$ is not on the unit circle so it's not necessary to explicitly exclude it if the locus was just the unit circle)

I'll post the solution tomorrow if nobody gets the right answer.

altSwift · Feb 10, 2018

Re: HSC 2018 MX2 Marathon

fan96 said:
As I said above, that's not the correct answer.

The locus is not $x^2 + y^2 = 1$ .

There are certain values of $z$ (specifically $x$ ) that need to be considered.

(By the way, $(0,0)$ is not on the unit circle so it's not necessary to explicitly exclude it if the locus was just the unit circle)

I'll post the solution tomorrow if nobody can figure it out.

rip, I'll have another go tonight

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