ATAR Notes: Forum

VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: itolduso on October 25, 2010, 10:19:49 pm

Title: i
Post by: itolduso on October 25, 2010, 10:19:49 pm: i^2 = -1
i = ?
sqrt(-1) = ?
Title: Re: i
Post by: brightsky on October 25, 2010, 10:40:45 pm: i = sqrt(-1)
Title: Re: i
Post by: m@tty on October 25, 2010, 10:48:33 pm: $e^{\pi i}=-1$

$\implies \sqrt{-1}=e^{\frac{\pi i}{2}$

$\implies i=\frac{1}{\pi}\log_e(-1)$
Title: Re: i
Post by: itolduso on October 25, 2010, 10:58:11 pm: what is ln(-1)/pi?
Title: Re: i
Post by: TrueTears on October 25, 2010, 11:00:53 pm: i

it's just definitions :D
Title: Re: i
Post by: m@tty on October 25, 2010, 11:04:28 pm: Quote from: itolduso on October 25, 2010, 10:58:11 pm
what is ln(-1)/pi?

Oh and realise that this is all imaginary, hence, don't expect to have a conceptual understanding of i or expect to be able to visualise where it would be on a number line...
Title: Re: i
Post by: itolduso on October 25, 2010, 11:06:31 pm: Quote from: TrueTears on October 25, 2010, 11:00:53 pm
i

it's just definitions :D

What definition are you referring to?
Title: Re: i
Post by: TrueTears on October 25, 2010, 11:07:01 pm: sqrt(-1) = i

in fact the proper definition is i^2 = -1 and sqrt(-1) = i is a consequence
Title: Re: i
Post by: itolduso on October 25, 2010, 11:11:56 pm: I havent seen such a definition. Where can i find it?
sqrt(-1) = i,
.: ln(-1)/pi = e^(ipi/2)?
Title: Re: i
Post by: Martoman on October 25, 2010, 11:41:26 pm: Quote from: m@tty on October 25, 2010, 11:04:28 pm
Quote from: itolduso on October 25, 2010, 10:58:11 pm
what is ln(-1)/pi?

Oh and realise that this is all imaginary, hence, don't expect to have a conceptual understanding of i or expect to be able to visualise where it would be on a number line...

I see it on a number line as hovering above it. It adds a dimention to numbers.
Title: Re: i
Post by: brightsky on October 25, 2010, 11:43:47 pm: Quote from: itolduso on October 25, 2010, 11:11:56 pm
I havent seen such a definition. Where can i find it?
sqrt(-1) = i,
.: ln(-1)/pi = e^(ipi/2)?

Well it so happens that $e^{ix} = \cos(x) + i\sin(x)$ . Hence plugging $x = \pi$ in, you get $e^{\pi i} = -1$ .

And just as m@tty has done above, do some manipulation of this:

$\log_e(-1) = \pi i$

$i = \frac{1}{\pi} \log_e(-1)$
Title: Re: i
Post by: m@tty on October 25, 2010, 11:48:17 pm: Quote from: Martoman on October 25, 2010, 11:41:26 pm
Quote from: m@tty on October 25, 2010, 11:04:28 pm
Quote from: itolduso on October 25, 2010, 10:58:11 pm
what is ln(-1)/pi?

Oh and realise that this is all imaginary, hence, don't expect to have a conceptual understanding of i or expect to be able to visualise where it would be on a number line...

I see it on a number line as hovering above it. It adds a dimention to numbers.

Yeah, so its not on the number line. It is separate, a new concept, part of the unknown...

On an Argand diagram you just treat it like you do the vertical axis on the Cartesian plane
Title: Re: i
Post by: TrueTears on October 25, 2010, 11:50:13 pm: Quote from: Martoman on October 25, 2010, 11:41:26 pm
Quote from: m@tty on October 25, 2010, 11:04:28 pm
Quote from: itolduso on October 25, 2010, 10:58:11 pm
what is ln(-1)/pi?

Oh and realise that this is all imaginary, hence, don't expect to have a conceptual understanding of i or expect to be able to visualise where it would be on a number line...

I see it on a number line as hovering above it. It adds a dimention to numbers.
yeah there actually IS a number line for complex numbers (check the wiki link below), in fact our real number line can not represent i, so i is kinda like the imaginary unit for the complex number system, you can think of it as the complex equivalent of a real number line.

Quote from: itolduso on October 25, 2010, 11:11:56 pm
I havent seen such a definition. Where can i find it?
sqrt(-1) = i,
.: ln(-1)/pi = e^(ipi/2)?
i dono, elementary algebra textbooks? or just read here

http://en.wikipedia.org/wiki/Imaginary_unit

i forgot the mathematician who formally defined i, maybe it says somewhere in the wiki article.
Title: Re: i
Post by: kyzoo on October 25, 2010, 11:53:21 pm: Quote from: Martoman on October 25, 2010, 11:41:26 pm
Quote from: m@tty on October 25, 2010, 11:04:28 pm
Quote from: itolduso on October 25, 2010, 10:58:11 pm
what is ln(-1)/pi?

Oh and realise that this is all imaginary, hence, don't expect to have a conceptual understanding of i or expect to be able to visualise where it would be on a number line...

I see it on a number line as hovering above it. It adds a dimention to numbers.

Wait a minute...if "i" adds a 2nd dimension to the numbers, then that means there's a 3rd dimenion, and a 4th, and so on?
Title: Re: i
Post by: Martoman on October 25, 2010, 11:55:34 pm: In a sense it adds a a certain dimentionality to them yes.
Title: Re: i
Post by: TrueTears on October 25, 2010, 11:56:40 pm: i think what martoman meant is, i is basically an extension to the real number system, ie, the complex number system, it's what makes our number system "complete", not necessarily visualizing our number system as a "dimension" , although you could possibly interpret $\mathbb{C}^2$ as the complex equivalent of $\mathbb{R}^2$ , a complex subspace? iono not my area of maths haha
Title: Re: i
Post by: itolduso on October 26, 2010, 07:03:24 am: (1) ln(-1)/pi = e^(ipi/2)?
(2) i^2 = -1, i = +/- sqrt(-1)?
(3) sqrt(-1) = +/- i?
Title: Re: i
Post by: Juddinator on October 26, 2010, 09:59:15 am: Hey guys,

We were given an example on how to find $z^5=1$ I was wondering if anyway could explain how $\frac{2\pi}{5}$ comes from in the following solution. I have looked through the textbook (Essentials) and I can't find it. It's the only part of the course I haven't gone over.

$z^5=1$
$Let z=r*cis(x)$
$\therefore r^5*cis(5x)=cis(\frac{\pi}{2})$
$\therefore r^5=1 and 5x=\frac{\pi}{2})$
$\therefore r=1 and x=\frac{\pi}{10})$
$\therefore r_1=cis(\frac{\pi}{10})$
$\therefore r_2=cis(\frac{\pi}{10}+\frac{2\pi}{5})$
$\therefore r_2=cis(\frac{\pi}{2})$
$\therefore r_3=cis(\frac{\pi}{2}+\frac{2\pi}{5})$
$\therefore r_3=cis(\frac{9\pi}{10})$
$\therefore r_4=cis(\frac{9\pi}{10}+\frac{2\pi}{5})$
$\therefore r_4=cis(\frac{13\pi}{10})$
$\therefore r_5=cis(\frac{13\pi}{10}+\frac{2\pi}{5})$
$\therefore r_5=cis(\frac{17\pi}{10})$

Cheers
Title: Re: i
Post by: superflya on October 26, 2010, 10:07:03 am: cis( $\theta$ + 2k $\pi$ )

both ur theta and 2k $\pi$ are multiplied by 1/5 then k values are subbed in to yield solutions.
Title: Re: i
Post by: Juddinator on October 26, 2010, 10:18:13 am: thanks superflya!
Title: Re: i
Post by: brightsky on October 26, 2010, 05:20:37 pm: Quote from: TrueTears on October 25, 2010, 11:56:40 pm
i think what martoman meant is, i is basically an extension to the real number system, ie, the complex number system, it's what makes our number system "complete", not necessarily visualizing our number system as a "dimension" , although you could possibly interpret $\mathbb{C}^2$ as the complex equivalent of $\mathbb{R}^2$ , a complex subspace? iono not my area of maths haha

Yeah basically the real number system can be visualised as a horizontal line extending to infinity both ways. Any real number lies somewhere on this graph. It can be thought of like the "x axis" of a cartesian graph. The complex number system is like an added "y-axis", so that now numbers can be placed anywhere on the plane.
Title: Re: i
Post by: pi on October 27, 2010, 06:28:36 pm: we proved a couple of weeks ago in GMA that i^i = e^((-pi/2)-(2npi)) -great 'fun', using Euler's formula (not sure if its on the GMA course though)

A few people in my class were really confused that a number that doesn't exist, when put to the power of a number that doesn't exist, is equal to a number that does exist... :o