HSC 2016 MX2 Marathon (archive) (1 Viewer)

leehuan · Nov 24, 2015

Re: HSC 2016 4U Marathon

When I did mechanics, and I had to resolve forces in upwards motion, I used -g.

When I had downwards motion, I took +g.

That is all I will say

InteGrand · Nov 24, 2015

Re: HSC 2016 4U Marathon

leehuan said:
When I did mechanics, and I had to resolve forces in upwards motion, I used -g.

When I had downwards motion, I took +g.

That is all I will say

$This is simply due to the choice of axis (there is only one axis since it is 1-dimensional motion). For upwards motion, it is generally more convenient to have the positive direction be upwards, so the weight force would point in the negative direction (as it points down). This would make the weight force vector be $\overrightarrow {W} = -mg$, where $g$ is a \textsl{positive} quantity (that's why we need the minus sign here).$

$For downwards motion, having the positive direction as downwards is more intuitive, so in this case the weight force points in the positive direction (down), and so the weight vector is $\overrightarrow{W} = mg$, where $g$ is a positive quantity.$

InteGrand · Nov 24, 2015

Re: HSC 2016 4U Marathon

$\textbf{NEW QUESTION}$

$Sketch and describe the locus $\arg (z) -\arg \left(z +i\right) = \frac{5\pi}{4}$.$

Ambility · Nov 24, 2015

Re: HSC 2016 4U Marathon

InteGrand said:
$\textbf{NEW QUESTION}$

$Sketch and describe the locus $\arg (z) -\arg \left(z +i\right) = \frac{5\pi}{4}$.$

I'm 90% sure this is wrong. If you could tell me what's wrong about it, that would be great. I can't even picture this locus in my head. It's a weird one.

$arg(z)-arg(z+i)=\frac{5\pi}{4}\\\tan^{-1}(\frac{y}{x})-\tan^{-1}(\frac{y+1}{x}) = \frac{5\pi}{4}\\\frac{y}{x}-\frac{y+1}{x}=1\\y-y-1=x\\x=-1$

leehuan · Nov 24, 2015

Re: HSC 2016 4U Marathon

Ambility said:
I'm 90% sure this is wrong. If you could tell me what's wrong about it, that would be great. I can't even picture this locus in my head. It's a weird one.

$arg(z)-arg(z+i)=\frac{5\pi}{4}\\\tan^{-1}(\frac{y}{x})-\tan^{-1}(\frac{y+1}{x}) = \frac{5\pi}{4}\\\frac{y}{x}-\frac{y+1}{x}=1\\y-y-1=x\\x=-1$

Line 3: You didn't use compound angles.

But wait for some experienced person to explain to you THAT locus

InteGrand · Nov 24, 2015

Re: HSC 2016 4U Marathon

Ambility said:
I'm 90% sure this is wrong. If you could tell me what's wrong about it, that would be great. I can't even picture this locus in my head. It's a weird one.

$arg(z)-arg(z+i)=\frac{5\pi}{4}\\\tan^{-1}(\frac{y}{x})-\tan^{-1}(\frac{y+1}{x}) = \frac{5\pi}{4}\\\frac{y}{x}-\frac{y+1}{x}=1\\y-y-1=x\\x=-1$

$Another problem is that in general, $\arg \left(\omega \right) \neq \tan^{-1} \left( \frac{y}{x}\right)$ (that equality is true if and only if $x>0$), where $x$ and $y$ are the real and imaginary parts of $\omega$, respectively. It is always true though that if $x\neq 0$, then $\tan \left(\arg \left(\omega\right) \right) =\frac{y}{x}$.$

kawaiipotato · Nov 25, 2015

Re: HSC 2016 4U Marathon

Ambility said:
I'm 90% sure this is wrong. If you could tell me what's wrong about it, that would be great. I can't even picture this locus in my head. It's a weird one.

$arg(z)-arg(z+i)=\frac{5\pi}{4}\\\tan^{-1}(\frac{y}{x})-\tan^{-1}(\frac{y+1}{x}) = \frac{5\pi}{4}\\\frac{y}{x}-\frac{y+1}{x}=1\\y-y-1=x\\x=-1$

The locus for the original question would be the major arc of a circle with centre to the left of the y axis with the angle between vectors z+i and z being 5pi/4

appleibeats · Nov 25, 2015

Re: HSC 2016 4U Marathon

determine the arcs specified by the following eqn. Sketch each one, showing the centre and radius of the associated circle.

arg ( z - 1 + i) / (z - 1 - i ) = pi/2

So i know it a circle. there are endpoints at (1, -1 ) and ( 1, 1) and that there are not included , hollow dot.

The angle between these two vectors when they meet is pi /2 .

The answers say it is a right hand side semicircle. Why is it a semicircle and not a full circle??

kawaiipotato · Nov 25, 2015

Re: HSC 2016 4U Marathon

appleibeats said:
determine the arcs specified by the following eqn. Sketch each one, showing the centre and radius of the associated circle.

arg ( z - 1 + i) / (z - 1 - i ) = pi/2

So i know it a circle. there are endpoints at (1, -1 ) and ( 1, 1) and that there are not included , hollow dot.

The angle between these two vectors when they meet is pi /2 .

The answers say it is a right hand side semicircle. Why is it a semicircle and not a full circle??

Draw up the vectors. Remember that anti-clockwise angles are positive. Drawing the left hand side semicircle and the vectors, the angle between them will be -pi/2 not pi/2 so you can only accept the right and side.

InteGrand · Nov 25, 2015

Re: HSC 2016 4U Marathon

appleibeats said:
determine the arcs specified by the following eqn. Sketch each one, showing the centre and radius of the associated circle.

arg ( z - 1 + i) / (z - 1 - i ) = pi/2

So i know it a circle. there are endpoints at (1, -1 ) and ( 1, 1) and that there are not included , hollow dot.

The angle between these two vectors when they meet is pi /2 .

The answers say it is a right hand side semicircle. Why is it a semicircle and not a full circle??

$The rule for sketching the locus of $\arg \left(\frac{z -z_1}{z -z_2}\right)= \alpha$ for a given $\alpha \in \left(0,\pi\right)$ is to start at the point $z_1$ and go counter-clockwise in a circular arc to the point $z_2$, with an angle of $\alpha$ being subtended in this arc (with open circles at $z_1$ and $z_2$, that is, these points are excluded from the locus). If you have the \textsl{other} arc, this corresponds to the locus $\arg \left(\frac{z -z_2}{z -z_1}\right)=\alpha$.$

InteGrand · Nov 25, 2015

Re: HSC 2016 4U Marathon

$\textbf{NEW QUESTION}$

$If $z$ is a complex number and $z\neq 0$, what does `$\arg (z)$' refer to? Does $\arg (z) = \arg (z) + 2\pi$? If so, does $0=2\pi$? Explain what is going on here.$

Drsoccerball · Nov 25, 2015

Re: HSC 2016 4U Marathon

Ambility · Nov 25, 2015

Re: HSC 2016 4U Marathon

InteGrand said:
$\textbf{NEW QUESTION}$

$If $z$ is a complex number and $z\neq 0$, what does `$\arg (z)$' refer to? Does $\arg (z) = \arg (z) + 2\pi$? If so, does $0=2\pi$? Explain what is going on here.$

arg(z) refers to the angle created by the line from the origin to the complex number z and the positive real axis on the Argand diagram. $arg(z)\neq arg(z)+2\pi$ as adding $2\pi$ to one side fundamentally changes the value. Take for example

$z=1+i\\arg(1+i)=\frac{\pi}{4}\\arg(z)+2\pi=\frac{\pi}{4}+2\pi=\frac{9\pi}{4}\\\frac{\pi}{4}\neq\frac{9\pi}{4}$

Adding $2\pi$ to the argument of a complex number will rotate the complex number a full revolution around the origin and will result in the same number. Finding the argument of a complex number and then adding $2\pi$ to that is not the same as the argument of the original complex number.

NEXT QUESTION: Using De Moivre's theorem, prove that $z^n+\frac{1}{z^n}=\cos n\theta$

Drsoccerball · Nov 25, 2015

Re: HSC 2016 4U Marathon

Ambility said:
NEXT QUESTION: Using De Moivre's theorem, prove that $z^n+\frac{1}{z^n}=\cos n\theta$

$z^n + \frac{1}{z^n} =2cos(n\theta)$

Ambility · Nov 25, 2015

Re: HSC 2016 4U Marathon

Drsoccerball said:
$z^n + \frac{1}{z^n} =2cos(n\theta)$

What?

leehuan · Nov 26, 2015

Re: HSC 2016 4U Marathon

Ambility said:
What?

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

InteGrand · Nov 26, 2015

Re: HSC 2016 4U Marathon

$\textbf{NEW QUESTION}$

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

appleibeats · Nov 26, 2015

Re: HSC 2016 4U Marathon

why is arg( (z)^1/2) = 1/2 arg z

leehuan · Nov 26, 2015

Re: HSC 2016 4U Marathon

appleibeats said:
why is arg( (z)^1/2) = 1/2 arg z

De Moivre's theorem states
When z=cos(x)+isin(x)
z^n = (cos(x)+isin(x))^n = cos(nx)+isin(nx)

arg(LHS) = arg(z^n)
Take note arg(z)=x
arg(RHS)=nx
So arg(z^n)=nx
arg(z^n)=n*arg(z)

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)

InteGrand · Nov 26, 2015

Re: HSC 2016 4U Marathon

appleibeats said:
why is arg( (z)^1/2) = 1/2 arg z

$Use De Moivre's Theorem. Note that in general powers of $z$ like this (i.e. non-integer powers) are multivalued and the idea of principal values of these is not in the syllabus as far as I'm aware. So $\frac{1}{2} \arg (z)$ would be the argument of one of the square roots of $z$; the other one will have argument $\pi + \frac{1}{2}\arg (z)$.$

HSC 2016 MX2 Marathon (archive) (1 Viewer)

Well-Known Member

Well-Known Member

Well-Known Member

Active Member

Well-Known Member

Well-Known Member

Well-Known Member

Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Attachments

Active Member

Well-Known Member

Active Member

Well-Known Member

Well-Known Member

Member

Well-Known Member

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)