Author Topic: 4U Maths Question Thread (Read 665235 times) Tweet Share

RuiAce · « **Reply #885 on:** February 26, 2017, 06:46:15 pm »

Quote from: VydekiE on February 26, 2017, 06:39:56 pm

Hi, it would be great if I could get some help on this question.
1.
a) If w is a seventh root of 1, w does not equal to 1, show that w^3+w^2+w+1+1/w+1/w^2+1/w^3=0
b) By letting z=w +1/w reduce this equation into a cubic equation in z.
Thank you!!

$\text{Note by the sum of a G.P.}\\ \begin{align*}1+\omega+\omega^2+\dots+\omega^6&= \frac{\omega^7-1}{\omega-1}\\ &= \frac{0}{\omega-1}\tag{*}\\ &= 0\end{align*}\\ \text{as }\omega^7 = 1\text{ from the question}$
$\text{So since }\omega^6+\omega^5+\dots+1=0\\ \text{Dividing by }\omega^3 (\text{ as obviously }\omega\neq 0)\\ \text{gives }\omega^3+\omega^2+\omega+\dots+\frac{1}{\omega^3}=0$
____________________________________
$\text{Let }z=\omega + \frac1\omega\text{ then by the binomial theorem}\\ \begin{align*}z^2&=\omega^2+2+\frac{1}{\omega^2}\\ z^3&=\omega^3+3\omega+\frac{3}{\omega}+\frac{1}{\omega^3}\end{align*}$
$\text{So by doing thorough rearranging to the result of part a)}\\ \begin{align*}\omega^3+\omega^2+\omega+1+\frac1\omega+\frac1{\omega^2}+\frac1{\omega^3}&=0\\ \left(\omega^3+3\omega+\frac{3}\omega+\frac{1}{\omega^3}\right)+\left(\omega^2+2+\frac{1}{\omega^2}\right)-2\left(\omega+\frac1\omega\right)-1&=0\\ z^3+z^2-2z-1&=0\end{align*}$

VydekiE · « **Reply #886 on:** February 26, 2017, 08:25:35 pm »

Thank you!!

michaelalt · « **Reply #887 on:** February 26, 2017, 10:03:25 pm »

Thanks Jake! Could you possibly help with this one as well? Its Q. 18 and I've drawn the diagram, but I don't know how to find the diagonal.

jakesilove · « **Reply #888 on:** February 26, 2017, 10:18:58 pm »

Quote from: michaelalt on February 26, 2017, 10:03:25 pm

Thanks Jake! Could you possibly help with this one as well? Its Q. 18 and I've drawn the diagram, but I don't know how to find the diagonal.

Hey! Weird question; you're not usually asked to find the cartesian equation of something on the argand diagram. Still, let's think about what we know.

Firstly, the line must be perpendicular to the diagonal AC. This is important, because we can find the gradient of the first diagonal, and use that to find the perpendicular gradient. Secondly, the line must bisect the diagonal AC. Now, we have a gradient and a point; all we need to find the equation!

Let's replace A and C with points on the cartesian plane; A(0, 3) and C(4, -5). The gradient of the line between them will be

$\frac{-5-3}{4-0}=-2$

So, the gradient of the perpendicular line will be 1/2.

Now, the midpoint of AC is (2, -1). So, the cartesian equation of the line perpendicular to the diagonal AC will be

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

So, we get the equation

$y=\frac{1}{2}x-2$

Great! That's the first part down. Now, the second point wants us to show that

$(1+2i)z+(1-2i)\bar{z}-8=0$

Let's let z=x+iy

$(1+2i)(x+iy)+(1-2i)(x-iy)-8=x+iy+2ix-2y+x-iy-2ix-2y-8=2x-4y-8=0$
$y=\frac{1}{2}x-2$

And we're done!

RuiAce · « **Reply #889 on:** February 26, 2017, 10:20:33 pm »

Quote from: michaelalt on February 26, 2017, 10:03:25 pm

Thanks Jake! Could you possibly help with this one as well? Its Q. 18 and I've drawn the diagram, but I don't know how to find the diagonal.

$\text{Recall that the diagonals of a rhombus}\\ \textit{bisect each other}\text{ AND }\textit{at right angles}.\\ \text{So the diagonals should be perpendicular to each other.}$
$\text{To cater for the bisection part}\\ \text{The midpoint of }AC\text{ is given by the complex number}\\ \frac{3i + (4-5i)}{2}=2-i\\ \text{so is the point }(2,-1)$
$\text{To cater for the perpendicularity part}\\ \text{The gradient of }AC\text{ is }m_{AC}=\frac{3+5}{0-4}=-2$
$\text{So since }BD\perp AC\text{ it has gradient }m=\frac12$
$\text{So by the point-gradient form, the equation of }BD\text{ is}\\ y-2 = \frac12(x+1)$
Woah. Rare instance that Jake beat me.

jakesilove · « **Reply #890 on:** February 26, 2017, 10:21:47 pm »

Quote from: RuiAce on February 26, 2017, 10:20:33 pm

$\text{Recall that the diagonals of a rhombus}\\ \textit{bisect each other}\text{ AND }\textit{at right angles}.\\ \text{So the diagonals should be perpendicular to each other.}$
$\text{To cater for the bisection part}\\ \text{The midpoint of }AC\text{ is given by the complex number}\\ \frac{3i + (4-5i)}{2}=2-i\\ \text{so is the point }(2,-1)$
$\text{To cater for the perpendicularity part}\\ \text{The gradient of }AC\text{ is }m_{AC}=\frac{3+5}{0-4}=-2$
$\text{So since }BD\perp AC\text{ it has gradient }m=\frac12$
$\text{So by the point-gradient form, the equation of }BD\text{ is}\\ y-2 = \frac12(x+1)$
Woah. Rare instance that Jake beat me.

Mate, I got one earlier today as well. Pick up your game son.

RuiAce · « **Reply #891 on:** February 26, 2017, 10:24:21 pm »

Quote from: jakesilove on February 26, 2017, 10:21:47 pm

Mate, I got one earlier today as well. Pick up your game son.

What did you want me to whilst I was dining out

jakesilove · « **Reply #892 on:** February 26, 2017, 10:26:07 pm »

Quote from: RuiAce on February 26, 2017, 10:24:21 pm

What did you want me to whilst I was dining out

Ninja answer a 4U question under the table, obviously.

RuiAce · « **Reply #893 on:** February 26, 2017, 10:29:04 pm »

Man... This guy...

Quote from: jakesilove on February 26, 2017, 10:26:07 pm

Ninja answer a 4U question under the table, obviously.

michaelalt · « **Reply #894 on:** February 27, 2017, 09:01:11 am »

Quote from: jakesilove on February 26, 2017, 10:18:58 pm

Hey! Weird question; you're not usually asked to find the cartesian equation of something on the argand diagram. Still, let's think about what we know.

Firstly, the line must be perpendicular to the diagonal AC. This is important, because we can find the gradient of the first diagonal, and use that to find the perpendicular gradient. Secondly, the line must bisect the diagonal AC. Now, we have a gradient and a point; all we need to find the equation!

Let's replace A and C with points on the cartesian plane; A(0, 3) and C(4, -5). The gradient of the line between them will be

$\frac{-5-3}{4-0}=-2$

So, the gradient of the perpendicular line will be 1/2.

Now, the midpoint of AC is (2, -1). So, the cartesian equation of the line perpendicular to the diagonal AC will be

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

So, we get the equation

$y=\frac{1}{2}x-2$

Great! That's the first part down. Now, the second point wants us to show that

$(1+2i)z+(1-2i)\bar{z}-8=0$

Let's let z=x+iy

$(1+2i)(x+iy)+(1-2i)(x-iy)-8=x+iy+2ix-2y+x-iy-2ix-2y-8=2x-4y-8=0$
$y=\frac{1}{2}x-2$

And we're done!

This makes perfect sense. Thank you so much! The only thing that confused me was when you substituted the gradient of the perpendicular line into the (y=mx+b) equation. You stated earlier that the gradient would be 1/2, however, when you substitute it into the (y=mx+b) equation, you've written it as -1/2.

Thank you so much otherwise Jake!

michaelalt · « **Reply #895 on:** February 27, 2017, 09:26:05 am »

Also if anyone could help me with this question.. Should I using De Moivre's theorem?

RuiAce · « **Reply #896 on:** February 27, 2017, 09:50:34 am »

Quote from: michaelalt on February 27, 2017, 09:01:11 am

This makes perfect sense. Thank you so much! The only thing that confused me was when you substituted the gradient of the perpendicular line into the (y=mx+b) equation. You stated earlier that the gradient would be 1/2, however, when you substitute it into the (y=mx+b) equation, you've written it as -1/2.

Thank you so much otherwise Jake!

Transcription error

Quote from: michaelalt on February 27, 2017, 09:26:05 am

Also if anyone could help me with this question.. Should I using De Moivre's theorem?

$\text{Combine De Moivre's with the difference of two squares expansion}\\ \begin{align*}[1+(\cos\theta+i\sin\theta)][1-(\cos\theta+i\sin\theta)]&=1-(\cos\theta+i\sin\theta)^2\\ &= 1-(\cos2\theta+i\sin2\theta)\end{align*}$

cutiepie30 · « **Reply #897 on:** February 27, 2017, 09:34:38 pm »

How would you do this question?

$\int_{0}^{\frac{\pi }{2}}\frac{1}{1+\tan (x)^{\sqrt{\pi }}} dx$

RuiAce · « **Reply #898 on:** February 27, 2017, 10:23:21 pm »

Quote from: cutiepie30 on February 27, 2017, 09:34:38 pm

How would you do this question?

$\int_{0}^{\frac{\pi }{2}}\frac{1}{1+\tan (x)^{\sqrt{\pi }}} dx$

$\text{Assuming that was meant to be }(\tan x)^{\sqrt \pi}$
$\text{Let }I=\int_{0}^{\frac{\pi }{2}}\frac{1}{1+(\tan x)^{\sqrt{\pi }}} dx\\ \text{Since }\tan x=\frac{\sin x}{\cos x}\text{, multiplying top}\\ \text{and bottom by }(\cos x)^{\sqrt \pi}\text{ gives}\\ I=\int_0^{\frac\pi{2}}\frac{(\cos x)^{\sqrt\pi}}{(\sin x)^{\sqrt\pi}+(\cos x)^{\sqrt\pi}}dx$
$\text{Call that equation 1}\\ \text{In equation 1, perform the trick }\int_0^a f(x)dx=\int_0^a f(a-x)dx\\ \text{with }a=\frac{\pi}{2}\\ \text{Then, noting the complementary identities e.g. }\sin \left(\frac{\pi}2-x\right)=\cos x\\ \text{We have }I=\int_0^{\frac\pi{2}}\frac{(\sin x)^{\sqrt\pi}}{(\sin x)^{\sqrt\pi}+(\cos x)^{\sqrt\pi}}dx\\ \text{Call that equation 2}$
$\text{Adding equations 2 and 1 tidies everything up}\\ \begin{align*}2I&=\int_0^{\frac\pi{2}}\frac{(\sin x)^{\sqrt\pi}+(\cos x)^{\sqrt\pi}}{(\sin x)^{\sqrt\pi}+(\cos x)^{\sqrt\pi}}dx\\ I&=\int_0^{\frac\pi 2}dx\\ &= \frac{\pi}{4}\end{align*}$
_____
$\text{In fact, it can be shown that for all real numbers }\alpha\\ \int_0^{\frac\pi 2}\frac{1}{1+\tan^\alpha x}dx=\frac{\pi}2$

cutiepie30 · « **Reply #899 on:** February 27, 2017, 11:51:30 pm »

Quote from: RuiAce on February 27, 2017, 10:23:21 pm

$\text{Assuming that was meant to be }(\tan x)^{\sqrt \pi}$
$\text{Let }I=\int_{0}^{\frac{\pi }{2}}\frac{1}{1+(\tan x)^{\sqrt{\pi }}} dx\\ \text{Since }\tan x=\frac{\sin x}{\cos x}\text{, multiplying top}\\ \text{and bottom by }(\cos x)^{\sqrt \pi}\text{ gives}\\ I=\int_0^{\frac\pi{2}}\frac{(\cos x)^{\sqrt\pi}}{(\sin x)^{\sqrt\pi}+(\cos x)^{\sqrt\pi}}dx$
$\text{Call that equation 1}\\ \text{In equation 1, perform the trick }\int_0^a f(x)dx=\int_0^a f(a-x)dx\\ \text{with }a=\frac{\pi}{2}\\ \text{Then, noting the complementary identities e.g. }\sin \left(\frac{\pi}2-x\right)=\cos x\\ \text{We have }I=\int_0^{\frac\pi{2}}\frac{(\sin x)^{\sqrt\pi}}{(\sin x)^{\sqrt\pi}+(\cos x)^{\sqrt\pi}}dx\\ \text{Call that equation 2}$
$\text{Adding equations 2 and 1 tidies everything up}\\ \begin{align*}2I&=\int_0^{\frac\pi{2}}\frac{(\sin x)^{\sqrt\pi}+(\cos x)^{\sqrt\pi}}{(\sin x)^{\sqrt\pi}+(\cos x)^{\sqrt\pi}}dx\\ I&=\int_0^{\frac\pi 2}dx\\ &= \frac{\pi}{4}\end{align*}$
_____
$\text{In fact, it can be shown that for all real numbers }\alpha\\ \int_0^{\frac\pi 2}\frac{1}{1+\tan^\alpha x}dx=\frac{\pi}2$

Thanks alot RuiAce

!!

i get how you get equation 1 and 2

but i dont get how you added equations 1 and 2 and simplified from there , could you please do those steps in more depth

Also for your last case is it meant to be pi/4 for all real number and not pi/2?

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