Author Topic: 4U Maths Question Thread (Read 665077 times) Tweet Share

RuiAce · « **Reply #825 on:** February 18, 2017, 06:42:44 pm »

Quote from: kinky_khan on February 18, 2017, 06:14:29 pm

HEYYYY Jake, I was struggling to answer the following questions and I was wondering if you could help me with them? thank you SOOOO MUCHH!!!

There are several questions here, and we will not do them all without seeing your attempts at them. I will provide the answer to a random choice of two.
$\text{Note that Q14 is not entirely accurate}\\ \text{It is possible for }\arg \frac{z}{w}=\pm\frac{\pi}{2}\\ \text{Proof by example: Let }z=1+i, \quad \omega = -1+i\\ \text{Then clearly }|z+\omega| = |z-\omega| = \sqrt{2}\\ \text{Yet }\arg \frac{z}{\omega}=-\frac{\pi}{2}$
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$\text{For the sake of illustration on the Argand diagram, we will assume }\omega\text{ is a constant}\\ \text{and that }z\text{ is a variable complex number, where }z=x+iy$
$\text{Suppose }\omega = a+bi\\ \text{so that the point }(a,b)\text{ on the Argand diagram corresponds to }\omega\\ \text{Then the locus is the perpendicular bisector of the line joining }(a,b)\text{ and }(-a,-b)$
GeoGebra simulation attached so you can play around with different values of a and b. This will be the diagram for reference.
A represents the complex number omega, and P represents any complex number z that satisfies the locus.
$\text{Playing around by moving }P\text{ in the GeoGebra simulation}\\ \text{verifies the result. To prove this:}$
$\text{First, note that }\\ \arg \frac{z}\omega = \arg z - \arg \omega\\ \text{Then, note from the diagram that}\\ \angle AOD = \arg \omega\quad \quad \angle POD = \arg z$
$\text{Substituting this in, it follows that}\\ \begin{align*}\arg \frac{z}{\omega} &= \arg z - \arg \omega\\ &= \angle POD - \angle AOD\\ &= \angle POA\end{align*}\\ \text{It now falls to find }\angle POA$
$\text{Note that in }\triangle POA\text{ and \triangle POB}\\ \begin{align*}PO&=PO\tag{common}\\ OA&=OB \tag{check with distance formula}\\ PA&=PB\tag{given!!!!!}\end{align*}\\ \therefore \triangle POA \equiv \triangle POB\, (SSS)$
$\text{You should now be able to use your diagram to show why }\angle POA = \frac{\pi}{2} (=90^\circ)$

RuiAce · « **Reply #826 on:** February 18, 2017, 07:03:28 pm »

Quote from: kinky_khan on February 18, 2017, 06:14:29 pm

HEYYYY Jake, I was struggling to answer the following questions and I was wondering if you could help me with them? thank you SOOOO MUCHH!!!

$\text{We follow the }\textbf{usual process}\text{ for 13 i)}\\ \begin{align*}\alpha^2+\beta^2+\gamma^2&=(\alpha+\beta+\gamma)^2-2(\alpha\beta+\alpha\gamma+\beta\gamma)\\ &= 0 -2 p\\ & = -2p\end{align*}$
$\text{As }\alpha \text{ IS a root, by substitution we obtain}\\ \alpha^3+p\alpha -1=0\\ \text{Multiply }\alpha\text{ to both sides of the equation to obtain}\\ \alpha^4+p\alpha^2-\alpha=0$
$\text{Similarly}\\ \beta^4+p\beta^2-\beta = 0\\ \gamma^4+p\gamma^2-\gamma = 0$
$\text{By adding these equations, and then substituting our earlier results}\\ \alpha^4+\beta^4+\gamma^4 + p(\alpha^2+\beta^2+\gamma^2)-(\alpha+\beta+\gamma)=0\\ \therefore \alpha^4+\beta^4+\gamma^4=2p^2$
$\text{Now, the fact that }\alpha^4+\beta^4+\gamma^4=2p^2\text{ doesn't really give the finishing touch}\\ \text{Rather, }\boxed{\alpha^2+\beta^2+\gamma^2=-2p}\text{ gives the finishing touch}$
$\text{Because }\alpha,\beta\text{ and }\gamma\text{ are all real, non-zero}\\ \text{we should have }\alpha^2 > 0\text{ and etc.}\\ \text{This is simply because the square of a number can never be negative}\\ \text{and only }0^2=0$
$\text{So if we add a ton of positive numbers, we should still get a positive number}\\ \text{Hence we have }\alpha^2+\beta^2+\gamma^2 > 0\\ \therefore -2p > 0 \implies p < 0\text{ as required.}$
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$\text{Note that the product of roots is }\alpha\beta\gamma = 1\\ \text{Hence, we have }\frac{\alpha}{\beta \gamma}=\frac{\alpha^2}{\alpha\beta\gamma}=\alpha^2\\ \text{so repeating the process, our roots are just }\alpha^2,\beta^2\text{ and }\gamma^2$
$\text{Follow the usual process - we replace }x\text{ with }\sqrt{x}\text{ to form our new polynomial}$
$\begin{align*}(\sqrt{x})^3+p\sqrt{x}-1&=0\\ \sqrt{x}(x+p)&=1\\ \implies x(x+p)^2&=1\\ x^3+2px^2+p^2x-1&=0\end{align*}$

hinakamishiro · « **Reply #827 on:** February 18, 2017, 07:49:05 pm »

Hi guys could i have some help with conics question 3 and 5b please? Thank you!

kiwiberry · « **Reply #828 on:** February 18, 2017, 08:40:51 pm »

Quote from: hinakamishiro on February 18, 2017, 07:49:05 pm

Hi guys could i have some help with conics question 3 and 5b please? Thank you!

Hey! for 5b):
$\begin{align*}\text{Length of PQ}&=|b\sin\theta-b\sin(-\theta)| \\ &=2b|\sin\theta| \\&=2b\sqrt{1-\cos^{2}x}\\&=2b\sqrt{1-e^{2}} \, (\text{from a)}) \\ \text{Since } b^{2}=a^{2}(1-e^{2}), PQ&=2b\times \frac{b}{a}=\frac{2b^{2}}{a}\end{align*}$

RuiAce · « **Reply #829 on:** February 18, 2017, 08:49:10 pm »

Quote from: hinakamishiro on February 18, 2017, 07:49:05 pm

Hi guys could i have some help with conics question 3 and 5b please? Thank you!

$\text{Note that }\\ \begin{align*}\cos(\pi + \theta)&=\cos \pi \cos \theta-\sin \pi \sin \theta=-\cos \theta\\\sin (\pi+\theta)&=\sin \pi\cos\theta + \cos \pi \sin \theta=-\sin \theta \end{align*}$
$\text{So }Q\text{ is literally just the point }(-a\cos \theta,-b\sin \theta)\\ \textbf{Trick: }\text{Notice how the }x\text{ and }y\text{ coordinates of }Q\\ \text{are really just the negatives of that of }P\\ \text{So you can show that }(0,0)\text{ is actually nothing more than the midpoint of }PQ$

michaelalt · « **Reply #830 on:** February 20, 2017, 08:59:59 am »

any help would be greatly appreciated

thanks

michaelalt · « **Reply #831 on:** February 20, 2017, 09:04:27 am »

I also don't understand how to do these questions (3 a & b).

jamonwindeyer · « **Reply #832 on:** February 20, 2017, 09:51:17 am »

Quote from: michaelalt on February 20, 2017, 08:59:59 am

any help would be greatly appreciated thanks

Hey hey! Here is the working for this question!

$\frac{iz}{z-1}=3+2i\\iz=(3+2i)(z-1)\\iz=(3+2i)z-3-2i\\(3+i)z=3+2i\\z=\frac{3+2i}{3+i}$

You can then manipulate that answer any way that suits you - I hope that helps!

I'll get to your next question in a little bit

michaelalt · « **Reply #833 on:** February 20, 2017, 10:56:57 am »

Quote from: jamonwindeyer on February 20, 2017, 09:51:17 am

Hey hey! Here is the working for this question!

$\frac{iz}{z-1}=3+2i\\iz=(3+2i)(z-i)\\iz=(3+2i)z+2-3i\\(3+i)z=-2+3i\\z=\frac{-2+3i}{3+i}$

You can then manipulate that answer any way that suits you - I hope that helps! I'll get to your next question in a little bit

Hey Jamon!
Thank you so much for this, if i was trying to manipulate it to get the answer 11/10 +3/10i how would you go about getting it?
Would i substitute the value of z into the original equation and work it out that way?

RuiAce · « **Reply #834 on:** February 20, 2017, 11:02:14 am »

Quote from: michaelalt on February 20, 2017, 10:54:36 am

Hey Jamon!
Thank you so much for this, if i was trying to manipulate it to get the answer 11/10 +3/10i how would you go about getting it?

$\text{Follow the }\textbf{usual process}\text{ of considering the complex conjugate}\\ \begin{align*}\frac{3+2i}{3+i}&= \frac{(3+2i)(3-i)}{(3+i)(3-i)}\\ &= \frac{9-3i+6i-2i^2}{9+1}\\ &=\frac{11}{10}+\frac{3i}{10} \end{align*}$

Jamon Edit - Fixed er up

RuiAce · « **Reply #835 on:** February 20, 2017, 11:04:51 am »

Quote from: michaelalt on February 20, 2017, 09:04:27 am

I also don't understand how to do these questions (3 a & b).

$\begin{align*}\frac{p+3qi}{2+i}&=\overline{1-i}\\ p+3qi&=(1+i)(2+i)\\ p+3qi&=1+3i\end{align*}$
$\text{So clearly, by equating the real and imaginary parts}\\ p=1\\ 3q=3$

jamonwindeyer · « **Reply #836 on:** February 20, 2017, 11:07:53 am »

Just came back to do those! Ta Rui

Edit: On the answer above, I wrote the question wrong in the second step: I'll go back and fix the errors in my working and then change Rui's to match, sorry!

michaelalt · « **Reply #837 on:** February 20, 2017, 12:01:49 pm »

Quote from: jamonwindeyer on February 20, 2017, 11:07:53 am

Just came back to do those! Ta Rui

Edit: On the answer above, I wrote the question wrong in the second step: I'll go back and fix the errors in my working and then change Rui's to match, sorry!

Thanks again Jamon + Rui! You guys are lifesavers.

michaelalt · « **Reply #838 on:** February 20, 2017, 12:25:10 pm »

Quote from: jamonwindeyer on February 20, 2017, 09:51:17 am

Hey hey! Here is the working for this question!

$\frac{iz}{z-1}=3+2i\\iz=(3+2i)(z-1)\\iz=(3+2i)z-3-2i\\(3+i)z=3+2i\\z=\frac{3+2i}{3+i}$

You can then manipulate that answer any way that suits you - I hope that helps! I'll get to your next question in a little bit

Can i just ask- how did you get to this (iz=(3+2i)z-3-2i) step? I don't understand the continuation from the previous step to this one.

michaelalt · « **Reply #839 on:** February 20, 2017, 12:41:51 pm »

Quote from: RuiAce on February 20, 2017, 11:04:51 am

$\begin{align*}\frac{p+3qi}{2+i}&=\overline{1-i}\\ p+3qi&=(1+i)(2+i)\\ p+3qi&=1+3i\end{align*}$
$\text{So clearly, by equating the real and imaginary parts}\\ p=1\\ 3q=3$

This makes sense! However, I tried doing part (b.) and it didn't work out as well as I thought it would...

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