ATAR Notes: Forum

VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: Damo17 on January 01, 2009, 03:55:25 pm

Title: Factorising-complex numbers
Post by: Damo17 on January 01, 2009, 03:55:25 pm: Need help please!

Factorise $z^4+64$ over $R$

I can easily factorise it over C (the previous question). But I am not sure how to get the answer they get for this.
Title: Re: Factorising-complex numbers
Post by: shinny on January 01, 2009, 04:00:20 pm: From your previous factors over C, just multiply the conjugate pairs together to get 2 quadratic factors over R.

EDIT: Namely, these being $z^2 -4z+8$ and $z^2+4z+8$
Title: Re: Factorising-complex numbers
Post by: Damo17 on January 01, 2009, 04:15:59 pm: I'm still not sure how to do it.

I factorised it as a pair of quadratics factors in c.
$(z^2-8i)(z^2+8i)$

And put it into linear factors over c.
$(z-2i-2)(z+2i+2)(z-2i+2)(z+2i-2)$

What do I use to work it out over R?
Title: Re: Factorising-complex numbers
Post by: Damo17 on January 01, 2009, 04:24:06 pm: Oh, sorry, I get it now.

(z-2i-2)(z+2i-2) and (z+2i+2)(z-2i+2)

I just got what you meant by conjugate pairs.

Thanks.
Title: Re: Factorising-complex numbers
Post by: shinny on January 01, 2009, 04:31:45 pm: Here's a method if you want to do it from scratch. This was my actual thought process for it, hence the unconess of it. There's probably a MUCH better way of doing this but I've avoided using complex numbers altogether in this method to show some algebraic techniques that could even be used in methods (although I doubt you'll ever need to);
$z^4+64=(z^2)^2+8^2$ <---you should know you're after two quadratic factors since it should be obvious they're not linear, and hence, this step involving squares seems the logical way to go
$=(z^2+8)^2+a$ , where a is something in terms of z. The reason for the a is that $(z^2+8)^2$ isn't equal to what we had before. I totally changed it around to make it resemble a quadratic, and the a is in there to compensate for any changes. This step seems a bit far fetched, but I guess it just comes with experience.
Next, by expanding our new brackets, we get $z^4+64+16z^2$ . Hence, our a is $-16z^2$ to cancel out the term we added in.
Hence, $z^4+64=(z^2+8)^2 -16z^2=(z^2+8)^2 -(4z)^2=(z^2-4z+8)(z^2+4z+8)$
But yeh, it's obviously best to use complex numbers, find the linear factors, then multiply the conjugate pairs.

EDIT: Mistakes @_@
Title: Re: Factorising-complex numbers
Post by: Damo17 on January 01, 2009, 04:40:21 pm: Good way. I think I understand it. Which way would be better/quicker to do? Would many people do it the way you did?
Title: Re: Factorising-complex numbers
Post by: shinny on January 01, 2009, 04:42:10 pm: Don't think anyone would do it that way...even I wouldn't normally. Like I said, I only did so to avoid using any complex numbers just for the sake of it. I think the most common and fastest would be to get the complex linear factors and yeh, multiply the conjugates.
Title: Re: Factorising-complex numbers
Post by: shinny on January 01, 2009, 04:47:28 pm: Sorry check it again, made a massive mistake before.
Title: Re: Factorising-complex numbers
Post by: Damo17 on January 01, 2009, 04:51:11 pm: Quote from: shinny on January 01, 2009, 04:47:28 pm
Sorry check it again, made a massive mistake before.

Yeah, I just seen that. As I went to post it showed your new post.
Title: Re: Factorising-complex numbers
Post by: Mao on January 01, 2009, 05:53:08 pm: Quote from: shinny on January 01, 2009, 04:42:10 pm
Don't think anyone would do it that way...even I wouldn't normally. Like I said, I only did so to avoid using any complex numbers just for the sake of it. I think the most common and fastest would be to get the complex linear factors and yeh, multiply the conjugates.

there are quite a few people who use that method, including my MUEP tutor :)
that kind of 'completing the square' can be very useful/