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annitaz

Posted: Sat Mar 06, 2010 12:36 am Post subject: Complex numbers

Forum Freshman

Joined: 05 Jan 2010
Posts: 7

1.Express cos4Θ as powers of cosΘ using the fact that if z = cosΘ + i sinΘ then
1/z = cosΘ-i sin = z^-1 so z^n +1/z^n = cos n Θ and z^n=1/Z^n= 2i sin Θ

2.Determine all solutions of the equation z^4-16 = 0

Can u please help me?

Heinsbergrelatz

Posted: Sat Mar 06, 2010 1:17 am Post subject:

Forum Masters Degree

Joined: 01 Aug 2009
Posts: 598
Location: Singapore

i think i know the solution to the second part, but not the 1st part..

2) $z^{4}-16 = 0$
what you must do is to "factorize" thereby giving;

$(z^{2}-4)(z^{2}+4)=0$

you can again factorize the above one more time;

(z+2i)(z-2i)(z+2)(z-2)=0

where $i=\sqrt{1}$

therefore the solution is $z=\pm2i$ OR $\pm2$
_________________
------------------

$G_\mu_\nu + \Lambda g_\mu_\nu=\frac{8\pi G}{c^{4}} T_\mu_\nu$

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DrRocket

Posted: Sat Mar 06, 2010 7:00 am Post subject: Re: Complex numbers

Forum Radioactive Isotope

Joined: 12 Aug 2008
Posts: 3114

annitaz wrote:

$z = |z|e^{i \theta}=|z|(cos(\theta)+isin(\theta))$ for some $\theta$

So

$z^4=|z|^4e^{i4 \theta} = 16 = 16 e^{i 2n \pi}$ for some n

From this we see that |z| = 2

and $4\theta = 2n \pi$ where

can be any integer.

So, $\theta = \frac {n \pi}{2}$ Now $\theta$ is only determined modulo $2 \pi$ so the distinct solutions are

$0, \frac {\pi}{2}, \pi , \frac {3 \pi}{2}$

And the corresponding solutions of the origiinal equation are

z=2, z = 2i, z=-2, z=-2i

Heinsbergrelatz

Posted: Sat Mar 06, 2010 9:24 am Post subject:

Forum Masters Degree

Joined: 01 Aug 2009
Posts: 598
Location: Singapore

wait i think i have th solution to your first question;

first of all the first question takes alot of working so i might make some mistakes as im writing it down..

$cos4\theta$ can be also written as $cos4\theta + isin4\theta$

later we can just extract the Cosine part, so the Sine here is not a great problem.

cos4\theta + isin4\theta can be deduced in to $|cis\theta|^{4}$ using De Moivre's theorem. this can be now rewritten as;

$(cos\theta + isin\theta)^{4}$ ,,,,this part you have to expand which is VERY tedious and easy to make errors, but this is the deduction;

$(cos\theta + isin\theta)^{3} = cos^{3}\theta-(3cos\theta)(sin^{2}\theta)+(3isin\theta)(cos^{2}\theta)-isin^{3}\theta$

after this you multiply $cos\theta + isin\theta$ to the result above;

$((cos^{3}\theta)-(3cos\theta)(sin^{2}\theta)+(3isin\theta)(cos^{2}\theta)-(isin^{3}\theta))(cos\theta + isin\theta)$

= $cos^{4}\theta+sin^{4}\theta-(6sin^{2}\theta)(cos^{2}\theta)+(4isin\theta)(sin^{3}\theta)-(4isin^{3}\theta)(cos\theta)$

we can already see it in a form x+yi

;

$cos^{4}\theta+sin^{4}\theta-(6sin^{2}\theta)(cos^{2}\theta)+i((4sin\theta)(sin^{3}\theta)-(4sin^{3}\theta)(cos\theta))$

so we just extract; $cos^{4}\theta+sin^{4}\theta-(6sin^{2}\theta)(cos^{2}\theta)$ ---- this is equal to the $cos4\theta$

all we have to do is now to simplify the result;

$cos^{4}\theta+sin^{4}\theta-(6sin^{2}\theta)(cos^{2}\theta)$ gives;

$cos^{4}\theta + (1-cos^{2}\theta)(1-cos^{2}\theta)-6((1-cos^{2}\theta)(cos^{2}\theta))$ -----remember that $1=sin^{2}\theta+cos^{2}\theta$

which finally gives;

$cos4\theta=8cos^{4}\theta-8cos^{2}\theta+1$

Note: you can also deduce $sin4\theta$
_________________
------------------

$G_\mu_\nu + \Lambda g_\mu_\nu=\frac{8\pi G}{c^{4}} T_\mu_\nu$

-------------------

Last edited by Heinsbergrelatz on Sat Mar 06, 2010 9:44 am; edited 1 time in total

annitaz

Posted: Sat Mar 06, 2010 9:42 am Post subject:

Forum Freshman

Joined: 05 Jan 2010
Posts: 7

thank u

annitaz

Posted: Sat Mar 06, 2010 11:00 am Post subject:

Forum Freshman

Joined: 05 Jan 2010
Posts: 7

Now Θ is only determined modulo 2Π

What does this mean?

DrRocket

Posted: Sat Mar 06, 2010 4:03 pm Post subject:

Forum Radioactive Isotope

Joined: 12 Aug 2008
Posts: 3114

annitaz wrote:

Now Θ is only determined modulo 2Π

What does this mean?

It means that any two values for $\theta$ that differ by $2 \pi$ define the same complex number.

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