Author Topic: i (Read 1810 times) Tweet Share

TrueTears · « **Reply #15 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

itolduso · « **Reply #16 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?

Juddinator · « **Reply #17 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

superflya · « **Reply #18 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.

Juddinator · « **Reply #19 on:** October 26, 2010, 10:18:13 am »

thanks superflya!

brightsky · « **Reply #20 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.

pi · « **Reply #21 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...

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