HSC 2016 MX2 Marathon (archive) (3 Viewers)

InteGrand · Nov 26, 2015

Re: HSC 2016 4U Marathon

leehuan said:
That's just the case when n=1/2 (NOTE: N ACTUALLY DOES NOT HAVE TO BE AN INTEGER. It's only called De Moivre's theorem when we deal with positive integers that's all. the statement is actually true for all real n)

$Note however that if $z = r\, \mathrm{cis} (\theta)$ is a general non-zero complex number and $\alpha = \frac{1}{n}$ for some positive integer $n$ (where $r,\theta \in \mathbb{R}$), then $z^{\alpha}$ will be \textsl{multivalued}. This means that we need to assign a principal value to the symbol $z^\alpha$, like how for positive real numbers, we have assigned a principal value for $x^{\frac{1}{2}}$ (namely we take that symbol to mean the \textsl{positive} square root of $x$). In general, it is not true that $\mathcal {PV}\left( z^\alpha \right)$ (where $\mathcal{PV}$ stands for principal value) will be the same as $r^\alpha \mathrm{cis} \left (\alpha \theta\right)$. However, it is true that $r^\alpha \mathrm{cis} \left (\alpha \theta\right)$ will be one of the values of the multivalued $z^\alpha$ (in fact this also is true for general non-real complex exponents $\alpha$, but non-real exponents are definitely not in the syllabus).$

$So in general, if a complex number is raised to a non-integer power, the result is multivalued, but using De Moivre's Theorem always gets you one of the values.$

Ambility · Nov 26, 2015

Re: HSC 2016 4U Marathon

leehuan said:
Correcting your question. If my mental arithmetic is correct, the answer is actually 2 * cos(ntheta)

Also, something tells me you made an assumption that |z|=1

Yeah you're right, nvm.

appleibeats · Nov 28, 2015

Re: HSC 2016 4U Marathon

arg( (z-3)/(z+3) ) = +_ 5pi/3

What is the angle at the circumference. The answer say it is pi/3 , but why is that??

InteGrand · Nov 28, 2015

Re: HSC 2016 4U Marathon

appleibeats said:
arg( (z-3)/(z+3) ) = +_ 5pi/3

What is the angle at the circumference. The answer say it is pi/3 , but why is that??

$This is because an angle at the circumference of a circle must be between 0 and $\pi$. So if the absolute value of the R.H.S. given to you is not in this range, we need to \textsl{add a multiple of} $2\pi$ to it to get it into the range $(0,\pi)$.$

$The reason we can do this is that $\arg ( \cdot )$ is \textsl{multivalued} and takes on any value in $2\pi$ distances of a principal value in $(-\pi,\pi ]$ (e.g. $\arg (1+i) =\frac{\pi}{4}, \frac{\pi}{4} \pm 2\pi , \frac{\pi}{4} \pm 2\cdot (2\pi ), \frac{\pi}{4}\pm 3\cdot (2\pi ), \ldots$).$

$So in the case where it is $\frac{5 \pi}{3}$, since this is not in the range $(0,\pi )$, we subtract $2 \pi$ from it to get it to $-\frac{\pi}{3}$, and then we can invert the expression in the arg to negate this angle to get $\frac{\pi}{3}$, explaining why the angle between the vectors is $\frac{\pi}{3}$.$

$For the $-\frac{5\pi}{3}$, we add $2\pi$ to get $\frac{\pi}{3}$, and so we get a `flipped' arc compared to the previous case.$

appleibeats · Nov 28, 2015

Re: HSC 2016 4U Marathon

Understand you clear explanation but just don't get this part:

we can invert the expression in the arg to negate this angle to get pi/3

InteGrand · Nov 28, 2015

Re: HSC 2016 4U Marathon

appleibeats said:
Understand you clear explanation but just don't get this part:

we can invert the expression in the arg to negate this angle to get pi/3

$This is based on the arg rule $\arg \left(\frac{1}{\omega} \right) = -\arg \left( \omega \right) $ (can also be written as $\arg \left( \omega \right) = -\arg\left( \frac{1}{\omega}\right)$. So for example, $\arg \left( 1+i\right) =-\arg \left(\frac{1}{1+i}\right)$.). So if we invert (take the reciprocal of) the thing inside an arg, we negate the value. So for the one where we had an arg of the quotient be $-\frac{\pi}{3}$, we can instead write the equation with a $+\frac{\pi}{3}$ on the right-hand side by inverting the quotient in the arg.$

Ambility · Dec 1, 2015

Re: HSC 2016 4U Marathon

InteGrand said:
$\textbf{NEW QUESTION}$

$Sketch (or for this Marathon, describe) the locus given by $\arg \left(z^3\right) = \frac{\pi}{3}$.$

$\\$Let $z=r(\cos\theta + i\sin\theta)\\arg(r^3\cos 3\theta+ r^3 i\sin 3\theta)=\frac{\pi}{3}\\3\theta = \frac{\pi}{3}\\\theta = \frac{\pi}{9}, \frac{7\pi}{9}, \frac{-5\pi}{9}\\$Three rays originating from the origin, travelling in the three arguments above. The centre is a hollow dot (undefined)$$

Ambility · Dec 1, 2015

Re: HSC 2016 4U Marathon

$$\textbf{NEW QUESTION}$\\\\$The complex numbers w and z with respective arguments $\phi$ and $\theta$ are two distinct points on the Argand diagram where $\phi < \theta$. The area of the triangle formed by the origin and these two points is A.$\\\\$Show that $z\bar{w}-w\bar{z}=4iA$

Drsoccerball · Dec 1, 2015

Re: HSC 2016 4U Marathon

Ambility said:
$$\textbf{NEW QUESTION}$\\\\$The complex numbers w and z with respective arguments $\phi$ and $\theta$ are two distinct points on the Argand diagram where $\phi < \theta$. The area of the triangle formed by the origin and these two points is A.$\\\\$Show that $z\bar{w}-w\bar{z}=4iA$

2013 HSC? Should I answer or let another 2016er answer?

leehuan · Dec 1, 2015

Re: HSC 2016 4U Marathon

Ambility said:
$$\textbf{NEW QUESTION}$\\\\$The complex numbers w and z with respective arguments $\phi$ and $\theta$ are two distinct points on the Argand diagram where $\phi < \theta$. The area of the triangle formed by the origin and these two points is A.$\\\\$Show that $z\bar{w}-w\bar{z}=4iA$

This question is definitely familiar. Probably what Drsoccerball said, but regardless of what year let a 16er do it.

InteGrand · Dec 1, 2015

Re: HSC 2016 4U Marathon

That question came with a diagram in the HSC paper if I recall correctly.

leehuan · Dec 1, 2015

Re: HSC 2016 4U Marathon

InteGrand said:
That question came with a diagram in the HSC paper if I recall correctly.

True. Because I remember so as well.

InteGrand · Dec 1, 2015

Re: HSC 2016 4U Marathon

$\textbf{UNANSWERED QUESTION}$

$\noindent Sketch (or for this Marathon, describe) the locus given by $\arg \left(z^3\right) = \frac{\pi}{3}$.$

Ambility · Dec 1, 2015

Re: HSC 2016 4U Marathon

InteGrand said:
$\textbf{UNANSWERED QUESTION}$

$\noindent Sketch (or for this Marathon, describe) the locus given by $\arg \left(z^3\right) = \frac{\pi}{3}$.$

Did I not answer it correctly above?

Ambility · Dec 1, 2015

Re: HSC 2016 4U Marathon

InteGrand said:
That question came with a diagram in the HSC paper if I recall correctly.

Yeah it did, I thought I'd include it since it is a little tricky.

InteGrand · Dec 1, 2015

Re: HSC 2016 4U Marathon

Ambility said:
Did I not answer it correctly above?

Oh sorry, didn't see you answered it. Yes, you answered it correctly.

leehuan · Dec 1, 2015

Re: HSC 2016 4U Marathon

Ambility said:
Yeah it did, I thought I'd include it since it is a little tricky.

The moment the diagram gets put in it's easy. So, good choice.

Ambility · Dec 2, 2015

Re: HSC 2016 4U Marathon

leehuan said:
The moment the diagram gets put in it's easy. So, good choice.

Haha, I guess the diagram does make it a bit easier to picture. If you're going to attempt the above one, I strongly recommend sketching a diagram.

Paradoxica · Dec 5, 2015

Re: HSC 2016 4U Marathon

$$\noindent Using induction, prove $ 42 | n^7 - n \; \; \forall n \in \mathbb N$

leehuan · Dec 5, 2015

Re: HSC 2016 4U Marathon

Never seen the vertical line before. What does it mean?

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