Author Topic: VCE Specialist 3/4 Question Thread! (Read 2548348 times) Tweet Share

Eugenet17 · « **Reply #2820 on:** January 07, 2014, 08:52:25 pm »

Quote from: abcdqd on January 07, 2014, 08:30:15 pm

Use De Moivre's thereom:
$\frac{1}{z}=z^{-1}=(\textup{cis}\theta)^{-1}=\textup{cis}(- \theta)$

We can re-write $-1 * 3\textup{cis}(\frac{2 \pi}{3})$ as $\textup{cis}(\pi)*3\textup{cis}(\frac{2\pi}{3})$ , which, using polar form multiplication properties, is equal to $3\textup{cis}(\frac{5\pi}{3})=3\textup{cis}(\frac{-\pi}{3})$ .
Now, since we are reflecting the complex number in the x-axis when taking its conjugate, we simply place a negative in front of the principle argument.
So the final answer is $3\textup{cis}(\frac{\pi}{3})$

Thank you so much!

Eugenet17 · « **Reply #2821 on:** January 08, 2014, 05:12:31 pm »

Im getting the Argument wrong for these two questions and im not sure why:
$i(\sqrt{3}-i)^7$

$\frac{(1+\sqrt{3}i)^3}{i(1-i)^5}$

brightsky · « **Reply #2822 on:** January 08, 2014, 05:17:33 pm »

i (sqrt(3) - i)^7 = cis(pi/2) (2cis(-pi/6))^7 = 2^7 cis(pi/2) cis(-7pi/6) = 2^7 cis(-2pi/3)

(1+sqrt(3)i)^3/(i(1-i)^5) = (2cis(pi/3))^3/[cis(pi/2)(sqrt(2) cis(-pi/4))^5] = 8 cis(pi)/[4sqrt(2) cis(-3pi/4)] = 2sqrt(2) cis(-pi/4)

hopefully no errors...

Eugenet17 · « **Reply #2823 on:** January 08, 2014, 05:27:12 pm »

Quote from: brightsky on January 08, 2014, 05:17:33 pm

i (sqrt(3) - i)^7 = cis(pi/2) (2cis(-pi/6))^7 = 2^7 cis(pi/2) cis(-7pi/6) = 2^7 cis(-2pi/3)

(1+sqrt(3)i)^3/(i(1-i)^5) = (2cis(pi/3))^3/[cis(pi/2)(sqrt(2) cis(-pi/4))^5] = 8 cis(pi)/[4sqrt(2) cis(-3pi/4)] = 2sqrt(2) cis(-pi/4)

hopefully no errors...

Ah that's right! Silly mistakes on my part, thanks

Eugenet17 · « **Reply #2824 on:** January 11, 2014, 01:15:43 pm »

How do i solve z^4-2z^2+4=0 in polar form?

Stick · « **Reply #2825 on:** January 11, 2014, 02:47:53 pm »

There's a couple of ways you could do it. This is how I'd tackle it:

Look if you get stuck

$Let$ $y=z^2$

$\therefore y^2-2y+4=0$

$y^2-2y+1-1+4=0$ (completing the square)

$(y-1)^2+3=0$ (factorising)

$(y-1)^2-(\sqrt{3}i)^2=0$ (employing the difference of perfect squares method)

$(y-1-\sqrt{3}i)(y-1+\sqrt{3}i)=0$ (factorising)

$\therefore y=1+\sqrt{3}i$ $or$ $y=1-\sqrt{3}i$

$\therefore z^2=1+\sqrt{3}i$ $or$ $z^2=1-\sqrt{3}i$

$\therefore r^2cis(2\theta)=1+\sqrt{3}i$ $or$ $r^2cis(2\theta)=1-\sqrt{3}i$ (employing De Moivre's theorem)

$r^2cis(2\theta)=2cis(\frac{\pi}{3})$ $or$ $r^2cis(2\theta)=2cis(-\frac{\pi}{3})$ (converting to polar form)

$\therefore r^2=2$ (equating both sides)

$\therefore r=\sqrt{2}$

$Consider$ $2cis(\frac{\pi}{3})$

$\therefore 2\theta+2n\pi=\frac{\pi}{3},$ $n \in Z$ (equating both sides)

$2\theta=\frac{\pi-6n\pi}{3},$ $n \in Z$

$\therefore \theta=\frac{\pi-6n\pi}{6},$ $n \in Z$

$Let$ $n=0,$ $\therefore \theta_1=\frac{\pi}{6}$

$Let$ $n=1,$ $\therefore \theta_2=\frac{-5\pi}{6}$

$Consider$ $2cis(-\frac{\pi}{3})$

$\therefore 2\theta+2n\pi=-\frac{\pi}{3},$ $n \in Z$ (equating both sides)

$2\theta=\frac{-\pi-6n\pi}{3},$ $n \in Z$

$\therefore \theta=\frac{-\pi-6n\pi}{6},$ $n \in Z$

$Let$ $n=0,$ $\therefore \theta_3=-\frac{\pi}{6}$

$Let$ $n=-1,$ $\therefore \theta_4=\frac{5\pi}{6}$

$Solutions:$

$\therefore z=\sqrt{2}cis(-\frac{5\pi}{6}),$ $z=\sqrt{2}cis(-\frac{\pi}{6}),$ $z=\sqrt{2}cis(\frac{\pi}{6}),$ $z=\sqrt{2}cis(\frac{5\pi}{6})$

I know that there's quite a few shortcuts that you can take in that method, but hopefully you'll understand it a lot better with the whole workings.

Eugenet17 · « **Reply #2826 on:** January 11, 2014, 06:23:42 pm »

Quote from: Stick on January 11, 2014, 02:47:53 pm

There's a couple of ways you could do it. This is how I'd tackle it:

Look if you get stuck
$Let$ $y=z^2$

$\therefore y^2-2y+4=0$

$y^2-2y+1-1+4=0$ (completing the square)

$(y-1)^2+3=0$ (factorising)

$(y-1)^2-(\sqrt{3}i)^2=0$ (employing the difference of perfect squares method)

$(y-1-\sqrt{3}i)(y-1+\sqrt{3}i)=0$ (factorising)

$\therefore y=1+\sqrt{3}i$ $or$ $y=1-\sqrt{3}i$

$\therefore z^2=1+\sqrt{3}i$ $or$ $z^2=1-\sqrt{3}i$

$\therefore r^2cis(2\theta)=1+\sqrt{3}i$ $or$ $r^2cis(2\theta)=1-\sqrt{3}i$ (employing De Moivre's theorem)

$r^2cis(2\theta)=2cis(\frac{\pi}{3})$ $or$ $r^2cis(2\theta)=2cis(-\frac{\pi}{3})$ (converting to polar form)

$\therefore r^2=2$ (equating both sides)

$\therefore r=\sqrt{2}$

$Consider$ $2cis(\frac{\pi}{3})$

$\therefore 2\theta+2n\pi=\frac{\pi}{3},$ $n \in Z$ (equating both sides)

$2\theta=\frac{\pi-6n\pi}{3},$ $n \in Z$

$\therefore \theta=\frac{\pi-6n\pi}{6},$ $n \in Z$

$Let$ $n=0,$ $\therefore \theta_1=\frac{\pi}{6}$

$Let$ $n=1,$ $\therefore \theta_2=\frac{-5\pi}{6}$

$Consider$ $2cis(-\frac{\pi}{3})$

$\therefore 2\theta+2n\pi=-\frac{\pi}{3},$ $n \in Z$ (equating both sides)

$2\theta=\frac{-\pi-6n\pi}{3},$ $n \in Z$

$\therefore \theta=\frac{-\pi-6n\pi}{6},$ $n \in Z$

$Let$ $n=0,$ $\therefore \theta_3=-\frac{\pi}{6}$

$Let$ $n=-1,$ $\therefore \theta_4=\frac{5\pi}{6}$

$Solutions:$

$\therefore z=\sqrt{2}cis(-\frac{5\pi}{6}),$ $z=\sqrt{2}cis(-\frac{\pi}{6}),$ $z=\sqrt{2}cis(\frac{\pi}{6}),$ $z=\sqrt{2}cis(\frac{5\pi}{6})$

I know that there's quite a few shortcuts that you can take in that method, but hopefully you'll understand it a lot better with the whole workings.

Thanks alot! I didn't think to let y=z^2 -.-

Stick · « **Reply #2827 on:** January 12, 2014, 10:52:18 am »

The key thing to remember is that since all the co-efficients were real numbers, all complex solutions occur in conjugate pairs. I've pretended to ignore that so that you can see the whole process, but that shortens the question down quite a bit.

lzxnl · « **Reply #2828 on:** January 12, 2014, 01:22:40 pm »

Here is my favourite way of doing the question.

$z^4 + 2z^2 + 4 = \frac {z^6-8}{z^2-2} = 0$
So set the numerator to zero while avoiding where the denom is zero. I trust this new equation is easier to solve.

How did I come up with this? By recognising the difference of two cubes formula and the similarities with the question. It's a dodgy method but I think it's cool.

Stick · « **Reply #2829 on:** January 12, 2014, 02:09:06 pm »

Quote from: lzxnl on January 12, 2014, 01:22:40 pm

Here is my favourite way of doing the question.

$z^4 + 2z^2 + 4 = \frac {z^6-8}{z^2-2} = 0$
So set the numerator to zero while avoiding where the denom is zero. I trust this new equation is easier to solve.

How did I come up with this? By recognising the difference of two cubes formula and the similarities with the question. It's a dodgy method but I think it's cool.

This is why this kid got a 49 in Specialist Maths.

That's a really cool method; thanks for sharing!

clıppy · « **Reply #2830 on:** January 12, 2014, 05:17:59 pm »

Bit stuck on a vectors question here, how do you show that two vectors intersect? I've attached the related question from the essentials textbook

brightsky · « **Reply #2831 on:** January 12, 2014, 05:26:12 pm »

Let X be the midpoint of DB and Y be the midpoint of CE. Find OX. Find OY. Observe that OX = OY. This proves both that the diagonals intersect and that the diagonals bisect each other. To find the acute angle between the two, use the dot product.

Eugenet17 · « **Reply #2832 on:** January 12, 2014, 05:36:38 pm »

brightsky · « **Reply #2833 on:** January 12, 2014, 05:51:24 pm »

'If and only if' problems normally warrant a proof with two parts.

Part 1. Prove that if 3 mod(z-1)^2 = mod(z+1)^2, then mod(z-2)^2 = 3.

3 (z-1)(conj(z) - 1) = (z+1)(conj(z) + 1)
3 (mod(z)^2 - z - conj(z) + 1) = mod(z)^2 + z + conj(z) + 1
3 mod(z)^2 - 3z - 3conj(z) + 3 - mod(z)^2 - z - conj(z) - 1 = 0
2 mod(z)^2 - 4z - 4conj(z) + 2 = 0
mod(z)^2 - 2z - 2conj(z) + 1 = 0
mod(z)^2 - 2z - 2conj(z) + 4 = 3
(z-2)(conj(z) - 2) = 3
mod(z-2)^2 = 3

Part 2. Prove that if mod(z-2)^2 = 3, then 3 mod(z-1)^2 = mod(z+1)^2.

It so happens that Part 2 of the problem has already been proven. We need only to reverse the process above to obtain the proof of Part 2.

Eugenet17 · « **Reply #2834 on:** January 12, 2014, 06:02:59 pm »

Quote from: brightsky on January 12, 2014, 05:51:24 pm

'If and only if' problems normally warrant a proof with two parts.

Part 1. Prove that if 3 mod(z-1)^2 = mod(z+1)^2, then mod(z-2)^2 = 3.

3 (z-1)(conj(z) - 1) = (z+1)(conj(z) + 1)
3 (mod(z)^2 - z - conj(z) + 1) = mod(z)^2 + z + conj(z) + 1
3 mod(z)^2 - 3z - 3conj(z) + 3 - mod(z)^2 - z - conj(z) - 1 = 0
2 mod(z)^2 - 4z - 4conj(z) + 2 = 0
mod(z)^2 - 2z - 2conj(z) + 1 = 0
mod(z)^2 - 2z - 2conj(z) + 4 = 3
(z-2)(conj(z) - 2) = 3
mod(z-2)^2 = 3

Part 2. Prove that if mod(z-2)^2 = 3, then 3 mod(z-1)^2 = mod(z+1)^2.

It so happens that Part 2 of the problem has already been proven. We need only to reverse the process above to obtain the proof of Part 2.

Wait does z x conj(z) = mod(z)^2?

Also, mod(z)^2 - 2z - 2conj(z) + 1 = 0
mod(z)^2 - 2z - 2conj(z) + 4 = 3

Where does the 3 come from?

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