# Thread: HSC 2018-2019 MX2 Marathon

1. ## Re: HSC 2018 MX2 Marathon

Originally Posted by sssona09
how about if re(z) is 2Im(z) and z^2-4i is real

I cant get the answer.. the answer is +- (2+i)
$\noindent Let z=x+yi, then \Re(z)=2\Im(z) \implies x=2y. Now z = 2y + yi \implies z^2 - 4i = 3y^2+(4y^2-4)i. Being real, 4y^2-4=0\implies y=\pm 1 \implies x = \pm 2. Thus, z=\pm (2+i)$

2. ## Re: HSC 2018 MX2 Marathon

Prove the ellipses:

$\frac{(x + k y)^2}{a^2} + \frac{y^2}{b^2} = 1, \frac{x^2}{a^2} + \frac{(kx + y)^2}{b^2} = 1$

where k is an arbitrary real number

have the same area as the ellipse

$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

3. ## Re: HSC 2018 MX2 Marathon

Originally Posted by Paradoxica
Prove the ellipses:

$\frac{(x + k y)^2}{a^2} + \frac{y^2}{b^2} = 1, \frac{x^2}{a^2} + \frac{(kx + y)^2}{b^2} = 1$

have the same area as the ellipse

$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$
Continuing on, prove that the area of the ellipse defined by

$\frac{(a x + b y)^2}{p^2} + \frac{(cx + dy)^2}{q^2} = 1$

has area

$\frac{\pi p q}{|ad - bc|}$

4. ## Re: HSC 2018 MX2 Marathon

$\\ Referring to the diagram above, where the point \ P \ is the intersection of curves \ y =x, \ y = \cos x, which region is larger in area, A_1 or A_2 ? Prove your answer without any use of a calculator (except for very basic facts like that \pi > 3 and so on, another exercise is to prove that \pi > 3)$

5. ## Re: HSC 2018 MX2 Marathon

Letting the value of the solution to x = cos(x) be r, the following area expressions are obtained through integration:

A₁ = sin(r) - r²/2

A₂ = 1 - sin(r) + r²/2

A₂ - A₁ = 1 - 2sin(r) + r²

r > sin(r) (proof is trivial and left as an exercise)

- 2sin(r) > - 2r

1 - 2sin(r) + r² > 1 - 2r + r² = (1-r)² > 0

A₂ - A₁ > 0

A₂ > A₁

7. ## Re: HSC 2018 MX2 Marathon

Originally Posted by Pakka
Change to polar form, apply de Moivre's theorem, done. For all of them.

8. ## Re: HSC 2018 MX2 Marathon

Could show me some working pls. Sorry for the trouble

9. ## Re: HSC 2018 MX2 Marathon

Never mind, I got it. Cheers, the advice made things easier.

10. ## Re: HSC 2018 MX2 Marathon

This one pls
Capture 3.JPG

11. ## Re: HSC 2018 MX2 Marathon

Originally Posted by Pakka
This one pls
Capture 3.JPG
Realise the denominator and use De Moivre's at the end to get the n in 2ntheta

12. ## Re: HSC 2018 MX2 Marathon

Appreciate some help with this: Find the range of values of |z| and arg(z) for |z-4-4i|=2sqrt(2)

13. ## Re: HSC 2018 MX2 Marathon

Originally Posted by CapitalSwine
Appreciate some help with this: Find the range of values of |z| and arg(z) for |z-4-4i|=2sqrt(2)
Sketch the locus of z and note that the modulus and argument of z use the join of the origin to a point on the circle.

14. ## Re: HSC 2018 MX2 Marathon

Originally Posted by 1729
Sketch the locus of z and note that the modulus and argument of z use the join of the origin to a point on the circle.
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

15. ## Re: HSC 2018 MX2 Marathon

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...

16. ## Re: HSC 2018 MX2 Marathon

Originally Posted by altSwift
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*}

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

17. ## Re: HSC 2018 MX2 Marathon

Therefore, using 3U method:

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

18. ## Re: HSC 2018 MX2 Marathon

Originally Posted by CapitalSwine
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 $\frac{\pi}{12} \leq \arg z \leq \frac{5\pi}{12}$

19. ## Re: HSC 2018 MX2 Marathon

Originally Posted by InteGrand
$\noindent \textbf{Hints.}$

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

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

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

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

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

$\noindent Therefore,$

\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

20. ## Re: HSC 2018 MX2 Marathon

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

21. ## 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...

22. ## Re: HSC 2018 MX2 Marathon

Originally Posted by fluffchuck
$\noindent After expanding the expression given by InteGrand, you should get an expression in terms of z^2, z^{-2}, z^4 and z^{-4}. From here, you should try to include (\star) somehow. To do this, you would need some terms in the form z^n+z^{-n}, such that n is an integer. From what we have, we can pair up z^2 and z^{-2} (this is n=2), then pair up z^4 and z^{-4} (this is n=4) etc.$
ah that makes more sense. we haven't done binomial expansions yet so looks like ill have to do it the long way haha

23. ## 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

24. ## Re: HSC 2018 MX2 Marathon

Originally Posted by si2136
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.

25. ## Re: HSC 2018 MX2 Marathon

Originally Posted by mrbunton
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?

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