Search results

$\noindent Try using properties of conjugation to help show that $w\overline{w} = 1$ (noting $z\bar{z} = 1$).$ $\noindent (The result won't generally hold if $a$ and $b$ are allowed to be non-real. It will though if $a$ and $b$ are real (and not both $0$ of course).)$

The tan function defined above is continuous but not monotonic. It's easy to come up with a discontinuous function that's not monotonic but is invertible. It suffices to take a graph that has two branches, one starting at the point (0, 2) and decreasing strictly, smoothly and asymptotically to...

Also, we don't need a function to be monotonic for it to be invertible, since we can easily form discontinuous functions that are not monotonic but pass the horizontal line test. However, you can prove as an exercise that a continuous one-to-one function must be monotonic.

$\noindent Yeah just monotonic isn't enough to imply invertibility, e.g. A constant function from $\mathbb{R}$ to $\mathbb{R}$ is monotonic but not invertible. `Strictly monotonic' would guarantee invertibility though (taking the codomain to be the range of the function).$

$\noindent It's because $z_{1}$ and $z_{2}$ have the same length (modulus), so $z_{1} + z_{2}$ forms the diagonal of a \emph{rhombus} (if they weren't the same length, it wouldn't be a rhombus, just a regular parallelogram. But since they're the same length, it's a special parallelogram, namely...

Re: HSC 2017 4U Marathon $\noindent Let $n = 2m$ be a positive even integer and let $x_{r} = \sin \left(\frac{\pi r^{2}}{2n}\right)$ for $r = 0,1,\ldots, n-1$. For integers $j\in \left\{0,1,\ldots, n-1\right\}$, define $\xi_{j} = \sum_{r = 0}^{n-1}x_{r} \mathrm{cis}\left(-\frac{2\pi j...

Not sure why the solutions keep saying stuff once they've shown that (w^2)^3 is equal to 1. Once you show this, you are done. (In fact similar reasoning shows that if w is any n-th root of unity, then so is w^k for any integer k). Maybe they meant to ask something else than what they actually wrote.

Re: 1995 Complex Number Problem Because 7 is prime. $\noindent Also in general, for positive integers $j$ and integers $n \geq 2$, we have that $\mathrm{cis}\left(\frac{j}{n}\cdot (2\pi)\right)$ is a primitive $n$-th root of unity if and only if $j$ and $n$ are coprime.$

The 'relative gaps' in your marks matter, so the marks aren't completely irrelevant. See http://www.boardofstudies.nsw.edu.au/hsc-results/moderation.html for further details.

Re: 1995 Complex Number Problem $\noindent You meant to put an $i$ before the sin, right? In that case, let $\omega = \mathrm{cis}\left(\frac{4\pi}{7}\right)$, then $\omega$ is a seventh root of unity and $7$ is prime, so $\omega$ is automatically a \emph{primitive} seventh root of unity (all...

Re: First Year Linear Algebra Marathon In fact, the matrix A is a doubly stochastic (in fact, symmetric) Markov matrix. By inspection it is the transition matrix of an irreducible Markov chain, and thus this chain (being irreducible and finite) has a unique stationary distribution. As it is...

Carlsen is World Champion 2016.

Jacaranda textbook.

With which part? Drongoski's first step was to multiply both sides of the inductive hypothesis by 12.

$\noindent The parabola has a vertical axis and vertex at the origin, so its equation is $y = ax^{2}$ for some $a$. We just need to find $a$. Since the line $y = 3x + \frac{3}{4}$ is tangent to the parabola, we must have that the quadratic $ax^{2} - 3x - \frac{3}{4}$ has discriminant $0$...

Re: HSC 2016 MX2 Combinatorics Marathon Here's a link to the game for anyone who wants to play it and/or do the above problem: https://www.mathsisfun.com/games/towerofhanoi.html .

Re: How do I get the exact gradient, with rational denominator, of the normal of para $\noindent Another way is to just find the `slope' of the normal imagining $y$ to be the first (`independent') variable instead, and then just invert this in the end. So we want the slope of normal to $x...

You appear to have left out the case of 0 letters between N and T.

Yes

Re: First Year Linear Algebra Marathon $\noindent Let $d$ and $n$ be positive integers and define $f(0) = 2$, $f(d) = -1$, and $f(k) = 0$ for $k \neq 0,d$. Let $A$ be the $n\times n$ matrix with $ij$ entry $f\left(|i-j|\right)$. Find in terms of $d$ and $n$ the solution $\mathbf{x} =...