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May 04, 2024, 12:50:13 am

Author Topic: Factorising-complex numbers  (Read 929 times)  Share 

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Damo17

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Factorising-complex numbers
« on: January 01, 2009, 03:55:25 pm »
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Need help please!

Factorise over

I can easily factorise it over C (the previous question). But I am not sure how to get the answer they get for this.
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shinny

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Re: Factorising-complex numbers
« Reply #1 on: January 01, 2009, 04:00:20 pm »
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From your previous factors over C, just multiply the conjugate pairs together to get 2 quadratic factors over R.

EDIT: Namely, these being and
« Last Edit: January 01, 2009, 04:13:40 pm by shinny »
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Damo17

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Re: Factorising-complex numbers
« Reply #2 on: January 01, 2009, 04:15:59 pm »
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I'm still not sure how to do it.

I factorised it as a pair of quadratics factors in c.


And put it into linear factors over c.


What do I use to work it out over R?
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Damo17

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Re: Factorising-complex numbers
« Reply #3 on: January 01, 2009, 04:24:06 pm »
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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.
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shinny

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Re: Factorising-complex numbers
« Reply #4 on: January 01, 2009, 04:31:45 pm »
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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);
<---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
, where a is something in terms of z. The reason for the a is that 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 . Hence, our a is to cancel out the term we added in.
Hence,
But yeh, it's obviously best to use complex numbers, find the linear factors, then multiply the conjugate pairs.

EDIT: Mistakes @_@
« Last Edit: January 01, 2009, 04:47:17 pm by shinny »
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Damo17

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Re: Factorising-complex numbers
« Reply #5 on: January 01, 2009, 04:40:21 pm »
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Good way. I think I understand it. Which way would be better/quicker to do? Would many people do it the way you did?
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shinny

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Re: Factorising-complex numbers
« Reply #6 on: January 01, 2009, 04:42:10 pm »
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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.
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shinny

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Re: Factorising-complex numbers
« Reply #7 on: January 01, 2009, 04:47:28 pm »
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Sorry check it again, made a massive mistake before.
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Damo17

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Re: Factorising-complex numbers
« Reply #8 on: January 01, 2009, 04:51:11 pm »
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Sorry check it again, made a massive mistake before.

Yeah, I just seen that. As I went to post it showed your new post.
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Mao

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Re: Factorising-complex numbers
« Reply #9 on: January 01, 2009, 05:53:08 pm »
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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/
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