# Thread: HSC 2018-2019 MX2 Marathon

yup

2. ## 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

3. ## 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

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

4. ## Re: HSC 2018 MX2 Marathon

Originally Posted by fan96
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...

5. ## Re: HSC 2018 MX2 Marathon

Originally Posted by fan96
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

6. ## Re: HSC 2018 MX2 Marathon

Originally Posted by mrbunton
=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)

7. ## Re: HSC 2018 MX2 Marathon

Originally Posted by fan96
This is an interesting question I found in the SGS notes.

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

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

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

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

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

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

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

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

$x^2 + y^2 = 1$

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

8. ## Re: HSC 2018 MX2 Marathon

$From a ruse paper:$

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

9. ## Re: HSC 2018 MX2 Marathon

Originally Posted by altSwift
$Noting that z \neq 0$

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

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

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

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

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

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

$x^2 + y^2 = 1$

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

10. ## Re: HSC 2018 MX2 Marathon

Originally Posted by fan96
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

11. ## Re: HSC 2018 MX2 Marathon

Originally Posted by altSwift
$From a ruse paper:$

$If z = \cos\theta + i\sin\theta, prove that \frac{2}{1 + z} = 1 - i\tan\left(\frac{\theta}{2}\right)$
An exercise in Patel (I believe 4H, Q8) has a similar, but harder question.

$\frac{2}{1+z}$

$= \frac{2}{(\cos \theta + 1) + i \sin \theta} \times \frac{(\cos \theta + 1) - i \sin \theta}{(\cos \theta + 1) - i \sin \theta}$

$=\frac{2\cos\theta + 2-2i\sin\theta}{(\cos\theta+1)^2+\sin^2\theta}$

$=\frac{2\cos\theta + 2-2i\sin\theta}{\cos^2\theta+2\cos\theta+1+(1-\cos^2\theta)}$

$=\frac{2\cos\theta + 2-2i\sin\theta}{2\cos\theta+2}$

$=1 - i\frac{\sin\theta}{\cos\theta+1}$

$\text{Let}\,t=\tan\left(\frac{\theta}{2}\right)$

$=1 - i\frac{\frac{2t}{1+t^2}}{\frac{1-t^2}{1+t^2}+\frac{1+t^2}{1+t^2}}$

$=1 - i\frac{2t}{2}$

$=1 - i\tan\left(\frac{\theta}{2}\right),\,\theta \neq \pm\pi$

12. ## Re: HSC 2018 MX2 Marathon

For anyone interested, the Patel question I was talking about is:

$\text{Prove that}\,\, (1-\cos\theta + 2i\sin\theta)^{-1} = \frac{1-2i\cot\left(\frac{\theta}{2}\right)}{5+3\cos\theta }$

13. ## Re: HSC 2018 MX2 Marathon

Originally Posted by fan96
An exercise in Patel (I believe 4H, Q8) has a similar, but harder question.

$\frac{2}{1+z}$

$= \frac{2}{(\cos \theta + 1) + i \sin \theta} \times \frac{(\cos \theta + 1) - i \sin \theta}{(\cos \theta + 1) - i \sin \theta}$

$=\frac{2\cos\theta + 2-2i\sin\theta}{(\cos\theta+1)^2+\sin^2\theta}$

$=\frac{2\cos\theta + 2-2i\sin\theta}{\cos^2\theta+2\cos\theta+1+(1-\cos^2\theta)}$

$=\frac{2\cos\theta + 2-2i\sin\theta}{2\cos\theta+2}$

$=1 - i\frac{\sin\theta}{\cos\theta+1}$

$\text{Let}\,t=\tan\left(\frac{\theta}{2}\right)$

$=1 - i\frac{\frac{2t}{1+t^2}}{\frac{1-t^2}{1+t^2}+\frac{1+t^2}{1+t^2}}$

$=1 - i\frac{2t}{2}$

$=1 - i\tan\left(\frac{\theta}{2}\right),\,\theta \neq \pm\pi$
Nice! Here is my solution:

$\frac{2}{1+z}$

$= \frac{2}{(\cos \theta + 1) + i \sin \theta} \times \frac{(\cos \theta + 1) - i \sin \theta}{(\cos \theta + 1) - i \sin \theta}$

$=\frac{2\cos\theta + 2-2i\sin\theta}{(\cos\theta+1)^2+\sin^2\theta}$

$=\frac{2\cos\theta + 2-2i\sin\theta}{\cos^2\theta+2\cos\theta+1+(1-\cos^2\theta)}$

$=\frac{2\cos\theta + 2-2i\sin\theta}{2\cos\theta+2}$

$=1 - i\frac{\sin\theta}{\cos\theta+1}$

$Using the double angle formulae...$

$=1 -i\frac{2\sin(\frac{\theta}{2})\cos(\frac{\theta}{2 })}{2\cos^2(\frac{\theta}{2})-1+1}$

$=1 - i\frac{\sin(\frac{\theta}{2})}{\cos(\frac{\theta}{ 2})}$

$= 1 -i\tan\left(\frac{\theta}{2}\right)$

$Where \theta \neq \pm\pi (not gunna lie I completely forgot about this haha)$

14. ## Re: HSC 2018 MX2 Marathon

It definitely seems to be something that most people forget. For this type of question it tends to not matter (in terms of marks), but being able to spot when a variable's domain has been restricted has saved me a few marks in exams.

(This might help with the locus problem I posted before...)

15. ## Re: HSC 2018 MX2 Marathon

Originally Posted by fan96
For anyone interested, the Patel question I was talking about is:

$\text{Prove that}\,\, (1-\cos\theta + 2i\sin\theta)^{-1} = \frac{1-2i\cot\left(\frac{\theta}{2}\right)}{5+3\cos\theta }$
$This is probably more convoluted than using the t-method$

$LHS = \frac{1}{1-\cos\theta + 2i\sin\theta} \times \frac{1-cos\theta -2i\sin\theta}{1-cos\theta -2i\sin\theta}$

$= \frac{1-\cos\theta -2i\sin\theta}{1-2\cos\theta + cos^2\theta + 4\sin^2\theta}$

$Using \sin^2\theta = 1 -\cos^2\theta$

$=\frac{1-\cos\theta -2i\sin\theta}{5 - 3cos^2\theta -2\cos\theta}$

$=\frac{1-\cos\theta -2i\sin\theta}{(3\cos\theta + 5)(1-\cos\theta)}$

$Splitting fractions and applying double angle formulae...$

$= \frac{1}{3\cos\theta + 5} - \frac{4i\sin(\frac{\theta}{2})\cos(\frac{\theta}{2 })}{(3\cos\theta + 5)(2-2\cos^2(\frac{\theta}{2}))}$

$= \frac{1}{3\cos\theta + 5} - \frac{2i\sin(\frac{\theta}{2})\cos(\frac{\theta}{2 })}{(3\cos\theta + 5)(1-\cos^2(\frac{\theta}{2}))}$

$Again using \sin^2\theta = 1 -\cos^2\theta$

$= \frac{1}{3\cos\theta + 5} - \frac{2i\sin(\frac{\theta}{2})\cos(\frac{\theta}{2 })}{(3\cos\theta + 5)(\sin^2(\frac{\theta}{2}))}$

$= \frac{1}{3\cos\theta + 5} - \frac{2i\cot(\frac{\theta}{2}))}{3\cos\theta + 5}$

$= \frac{1-2i\cot(\frac{\theta}{2})}{5+3\cos\theta}$

$= RHS$

16. ## Re: HSC 2018 MX2 Marathon

Originally Posted by fan96
It definitely seems to be something that most people forget. For this type of question it tends to not matter (in terms of marks), but being able to spot when a variable's domain has been restricted has saved me a few marks in exams.

(This might help with the locus problem I posted before...)
Ooh these hints

17. ## Re: HSC 2018 MX2 Marathon

Originally Posted by fan96
This is an interesting question I found in the SGS notes.

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

$\mathrm{Re}\left(z-\frac{1}{z}\right) = 0$
 Spoiler (rollover to view): let z= rcosθ+isinθ z-1/z = (r-1/r)cosθ + i(r+1/r)sinθ Re(z-1/z) = (r-1/r)cosθ If LHS = 0, then (r-1/r) = 0 or cosθ = 0 hence, r=1 or θ= ±(2n+1)π/2 Hence the locus is x^2+y^2 = 1 and x = 0 (excluding (0,0))

18. ## Re: HSC 2018 MX2 Marathon

Originally Posted by jathu123
 Spoiler (rollover to view): let z= rcosθ+isinθ z-1/z = (r-1/r)cosθ + i(r+1/r)sinθ Re(z-1/z) = (r-1/r)cosθ If LHS = 0, then (r-1/r) = 0 or cosθ = 0 hence, r=1 or θ= ±(2n+1)π/2 Hence the locus is x^2+y^2 = 1 and x = 0 (excluding (0,0))
ahahaha I forgot to consider x = 0 jesus

19. ## Re: HSC 2018 MX2 Marathon

A classic one

$Simplify \sin(x)+\sin(2x) + \sin(3x) + ... + \sin(nx)$

20. ## Re: HSC 2018 MX2 Marathon

Originally Posted by jathu123
 Spoiler (rollover to view): let z= rcosθ+isinθ z-1/z = (r-1/r)cosθ + i(r+1/r)sinθ Re(z-1/z) = (r-1/r)cosθ If LHS = 0, then (r-1/r) = 0 or cosθ = 0 hence, r=1 or θ= ±(2n+1)π/2 Hence the locus is x^2+y^2 = 1 and x = 0 (excluding (0,0))
Yep.

If you used $z = x + iy$, you ended up with the locus

$x^3 + xy^2 = x$

It's tempting to divide both sides by $x$, but if you performed the division you were implicitly assuming that $x \neq 0$, which lost solutions.

Clearly if $x=0$, any value of $y$ is a solution, except for zero since $z \neq 0$.

That can also be shown by considering
$\text{Re}\left(iy + \frac{1}{iy}\right)= \text{Re}\left(iy - i\frac{1}{y}\right)$
which is zero for all $y \neq 0$.

To graph $x^3 + xy^2 = x$, you could consider two separate cases: $x=0$, and $x\neq 0$.

This is the solution from the SGS notes:

21. ## Re: HSC 2018 MX2 Marathon

Originally Posted by mrbunton
polynomial/complex number question:
cot(x)= y+1
a,b are roots of z^2+2z-1=0
prove that ( (y+a)^n - (y+b)^n ) / (a-b) = sin(nx) / sinx^n
I tried taking out a factor of

$\frac{1}{sin^n(x)}$

from the top two brackets but I still cant get to the LHS... any hints?

22. ## Re: HSC 2018 MX2 Marathon

Originally Posted by altSwift
I tried taking out a factor of

$\frac{1}{sin^n(x)}$

from the top two brackets but I still cant get to the LHS... any hints?
make sure to use all the given equations.
eg: a and b represent their own complex numbers (on the lhs), try finding them;
its easier approaching it from the lhs to the rhs;without trying to take out any common factors immediately

23. ## Re: HSC 2018 MX2 Marathon

Originally Posted by mrbunton
make sure to use all the given equations.
eg: a and b represent their own complex numbers (on the lhs), try finding them;
its easier approaching it from the lhs to the rhs;without trying to take out any common factors immediately
Are you sure you wrote the question right? Cause a and b are real numbers (-1+-√2) not complex numbers in the form a+ib (b=/=0) (assuming that's what you intended it to be)

24. ## Re: HSC 2018 MX2 Marathon

Originally Posted by mrbunton
make sure to use all the given equations.
eg: a and b represent their own complex numbers (on the lhs), try finding them;
its easier approaching it from the lhs to the rhs;without trying to take out any common factors immediately
I used all the stuff you gave me and went from lhs -> rhs but I end up with crappy trig inside the ()^n that I haven't been able to transform into cosx - isinx to use with de moivre

ill write it up in latex later this afternoon

25. ## Re: HSC 2018 MX2 Marathon

Originally Posted by jathu123
Are you sure you wrote the question right? Cause a and b are real numbers (-1+-√2) not complex numbers in the form a+ib (b=/=0) (assuming that's what you intended it to be)
Originally Posted by altSwift
I used all the stuff you gave me and went from lhs -> rhs but I end up with crappy trig inside the ()^n that I haven't been able to transform into cosx - isinx to use with de moivre

ill write it up in latex later this afternoon
soz dudes i made a typo; z^2-2z+2=0 where a and b are roots
i apologise for any inconvenience this may have caused

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