Author Topic: Complex Numbers (Read 21645 times) Tweet Share

uni09 · « **Reply #45 on:** April 02, 2009, 08:51:41 pm »

Quote from: d0minicz on April 02, 2009, 08:38:13 pm

in the back it says Cis Pi/2 , cis (-5pi/6) and cis (-pi/6) on the argand diagram i am confused.

we don't use cis round here these days

d0minicz · « **Reply #46 on:** April 02, 2009, 08:52:52 pm »

then wat

uni09 · « **Reply #47 on:** April 02, 2009, 08:56:51 pm »

TrueTears · « **Reply #48 on:** April 02, 2009, 09:00:45 pm »

Quote from: uni09 on April 02, 2009, 08:56:51 pm

$e^i^\theta$

lol nice. I type that on the calc, cos i don't know the Cis function button haha

d0minicz · « **Reply #49 on:** April 03, 2009, 09:31:56 pm »

Let $z= a+ ib$ where a and b are real numbers, and let $z^2 = 3+4i$ .
a) Find the equations in terms of a and b by equating real and imaginary parts. I got $a^2 - b^2 =3$ and $2ab = 4$ .

b) Find the values of a and b and hence the square roots of $3+4i<br />$ . - need help found a and b now how do i find the square roots of it ?

thanks

TonyHem · « **Reply #50 on:** April 03, 2009, 09:45:01 pm »

Not 100% sure.
Think its
$Z = \sqrt{3+4i}$
$Z^2 = 3+4i$
$a = 3, b= 4$
$x^2 = \frac{3+\sqrt{3^2+4^2}}{2}$
$x^2 = 4$
$x =\pm2$
$y^2 = x^2-a$
$y^2 = 4-3$
$y^2 = 1$
$y =\pm 1$
i think.. So roots are $z = 2+i$ and $Z= -2-i.$

------------------------------------
Edit:

$z = \sqrt{24+7i}$
$z^2 = 24+7i$
$a = 24, b = 7$

$x^2 = \frac{a+\sqrt{a^2+b^2}}{2}$
$x^2 = \frac{24+\sqrt{24^2+7^2}}{2}$
$x^2 = \frac{24+25}{2}$
$x^2 = \frac{49}{2}$
$x = \pm \frac{7\sqrt{2}}{2}$

$y^2 = x^2- a$
$y^2 = \frac{49}{2} -\frac{48}{2}$
$y^2 = 1/2$
$y =\pm\frac{\sqrt{2}}{2}$

$z = x+yi = \pm\frac{7\sqrt{2}}{2}+ \frac{\sqrt{2}}{2}i$
and take out $\frac{\sqrt{2}}{2}$

I spent 50 years on this because of my shitty latex skills - fixing all these dumb code things. ={, why must it be so difficult? :S

d0minicz · « **Reply #51 on:** April 03, 2009, 10:15:32 pm »

Find the square roots of $24 + 7i$ . just unsure of the quadratic i end up with and how to factorise it. thanks

Flaming_Arrow · « **Reply #52 on:** April 03, 2009, 10:34:55 pm »

$z= a + bi$ and $z^2+24+7i$

equate the real and the imaginery

$a^2-b^2=24 2ab=7$

$b=\frac{7}{2a}$

$a^2-\frac{49}{4a^2} = 24$

$a= \pm \frac{7}{2} \sqrt{2}$

sub that into $2ab=7$

which yields $b=\pm\frac{\sqrt{2}}{2}$

$\therefore \pm \frac{\sqrt{2}}{2} (7+i)$

d0minicz · « **Reply #53 on:** April 03, 2009, 11:00:55 pm »

can u show me the steps to solve for a plz
thanks =]

d0minicz · « **Reply #54 on:** April 04, 2009, 12:40:45 am »

Find the solutions of the equation $z^4 - 2z^2 + 4 = 0$ in polar form.

thankz

Over9000 · « **Reply #55 on:** April 04, 2009, 01:21:57 am »

let $w = z^2$

$w^2 - 2w + 4 = 0$

$w^2 - 2w + 1 - 1 + 4 =0$

$(w-1)^2 + 3= 0$

$(w-1)^2 - 3i^2 = 0$

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

$w = 1 - \sqrt{3} or 1 + \sqrt{3}$

now you can solve this for $z$ using de moives theroem.

sub $z^2 = w$ back in to yield $z^2 = 1 - \sqrt{3} or 1 + \sqrt{3}$

d0minicz · « **Reply #56 on:** April 04, 2009, 01:39:44 pm »

1. a) Find the square roots of $1+i$ by:
i) cartesian methods : i got $\pm (\sqrt \frac {1+\sqrt 2}{2} + \sqrt \frac {\sqrt 2 -1}{2} )$
ii) De Moivre's theorem : i got $z = 2^\frac {1}{4} cis (\frac {\pi}{8})$ , $z = 2^\frac {1}{4} cis (\frac {-7\pi}{8})$

b) Hence find exact values of $cos \frac {\pi}{8}$ and $sin \frac {\pi}{8}$ . - need help.

thanks

kamil9876 · « **Reply #57 on:** April 04, 2009, 02:04:57 pm »

from ii.):

$z=2^{\frac{1}{4}}(cos(\pi/8)) + i2^{\frac{1}{4}}(cos(\pi/8))$

Now you can equate the real parts of above expression with you're cartesian expression and hence you will get $cos(\pi/8)$ in terms of some surds and rationals. Same thing can be done with imaginary parts for $sin(\pi/8)$ or just use pythagoras theorem and your result from equating real parts.

d0minicz · « **Reply #58 on:** April 05, 2009, 05:10:57 pm »

Illustrate $2 Im(z) = Re(z)$ on an argand diagram.
I dont understand how to do these questions. thanks.

TonyHem · « **Reply #59 on:** April 05, 2009, 05:27:53 pm »

Imz
Z = x+yi
imaginary = y

U get re now right?

Search the forums now!

We have moved!

Author Topic: Complex Numbers (Read 21645 times) Tweet Share

uni09

Re: Complex Numbers

d0minicz

Re: Complex Numbers

uni09

Re: Complex Numbers

TrueTears

Re: Complex Numbers

d0minicz

Re: Complex Numbers

TonyHem

Re: Complex Numbers

d0minicz

Re: Complex Numbers

Flaming_Arrow

Re: Complex Numbers

d0minicz

Re: Complex Numbers

d0minicz

Re: Complex Numbers

Over9000

Re: Complex Numbers

d0minicz

Re: Complex Numbers

kamil9876

Re: Complex Numbers

d0minicz

Re: Complex Numbers

TonyHem

Re: Complex Numbers

Recent Posts

User:
Password: