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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: Matt The Rat on February 24, 2008, 04:27:08 pm

Title: Complex Numbers Question
Post by: Matt The Rat on February 24, 2008, 04:27:08 pm: Hey everyone,

I've come up with a question, from a mate's SAC from last year, which I can't figure out. Any help would be appreciated. :)

Q b) Let w be a complex root of unity. Prove that $w^2$ is also a complex cube root of unity.

c) Prove that $1 + w + w^2 =0$

Thanks in advance!
Title: Re: Complex Numbers Question
Post by: Neobeo on February 24, 2008, 04:44:10 pm: Second part:

$1+w+w^2=\frac{w^3-1}{w-1}=0$ , since $w^3=1$ and $w \neq 1$
Title: Re: Complex Numbers Question
Post by: Mao on February 24, 2008, 05:53:54 pm: if we express w in polar form

$w=cis(\theta)$

$w^3=cis(3 \theta)=cis(0 + 2 k \pi)$

$\theta=\frac{2 k \pi}{3}$

then w² can be expressed as $w^2=cis(2 \theta)=cis(\frac{4 k \pi}{3})$

and we can show that $\left( w^2 \right) ^3=cis(4 k \pi)=cis(0)=1$

therefore w² is also a complex cube root of unity, QED :D
Title: Re: Complex Numbers Question
Post by: Juddinator on June 20, 2010, 04:38:22 pm: Im having trouble getting the jist of these type of questions. Like I get somewhere, then fall over, then get up and go again but get the worng answer lol.

http://i46.tinypic.com/2n9mknd.jpg
Title: Re: Complex Numbers Question
Post by: Juddinator on June 20, 2010, 07:47:10 pm: hope the link works out alright....
Title: Re: Complex Numbers Question
Post by: m@tty on June 21, 2010, 05:32:31 pm: You should be able to get all of the part As. Use $\cos\left(\frac{\pi}{2}-\theta\right)=\sin(\theta)$ , $\sin\left(\frac{\pi}{2}-\theta\right)=\cos(\theta)$ , $\cos(-\theta)=\cos(\theta)$ and $\sin(-\theta)=-\sin(\theta)$ .

5. b. ii. $(\sin(\theta)+\cos(\theta)i)(\cos(\theta)+\sin(\theta)i)=cis\left(\frac{\pi}{2}-\theta\right)\cdot cis(\theta)$

Then, because $cis(a) \cdot cis(b)=cis(a+b)$

$cis\left(\frac{\pi}{2}-\theta\right)\cdot cis(\theta)=cis\left(\frac{\pi}{2}\right)=i$ .

6. b. i. $(\cos(\theta)-\sin(\theta)i)^5=\left(cis(-\theta)\right)^5=cis(-5\theta)$

7. b. iii. $(\sin(\theta)-\cos(\theta)i)^2(\cos(\theta)-\sin(\theta)i)=\left(cis\left(\theta - \frac{\pi}{2}\right)\right)^2cis(-\theta)$

$\left(cis\left(\theta - \frac{\pi}{2}\right)\right)^2cis(-\theta)=cis(2\theta-\pi)cis(-\theta)=cis(\theta-\pi)$

Do that kind of thing :)
Title: Re: Complex Numbers Question
Post by: Juddinator on June 21, 2010, 06:28:35 pm: Quote from: m@tty on June 21, 2010, 05:32:31 pm
You should be able to get all of the part As. Use $\cos\left(\frac{\pi}{2}-\theta\right)=\sin(\theta)$ , $\sin\left(\frac{\pi}{2}-\theta\right)=\cos(\theta)$ , $\cos(-\theta)=\cos(\theta)$ and $\sin(-\theta)=-\sin(\theta)$ .

5. b. ii. $(\sin(\theta)+\cos(\theta)i)(\cos(\theta)+\sin(\theta)i)=cis\left(\frac{\pi}{2}-\theta\right)\cdot cis(\theta)$

Then, because $cis(a) \cdot cis(b)=cis(a+b)$

$cis\left(\frac{\pi}{2}-\theta\right)\cdot cis(\theta)=cis\left(\frac{\pi}{2}\right)=i$ .

6. b. i. $(\cos(\theta)-\sin(\theta)i)^5=\left(cis(-\theta)\right)^5=cis(-5\theta)$

7. b. iii. $(\sin(\theta)-\cos(\theta)i)^2(\cos(\theta)-\sin(\theta)i)=\left(cis\left(\theta - \frac{\pi}{2}\right)\right)^2cis(-\theta)$

$\left(cis\left(\theta - \frac{\pi}{2}\right)\right)^2cis(-\theta)=cis(2\theta-\pi)cis(-\theta)=cis(\theta-\pi)$

Do that kind of thing :)
yeh, im clicking now, thanks man