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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: Becky2012 on May 03, 2012, 09:12:03 pm

Title: Complex Numbers
Post by: Becky2012 on May 03, 2012, 09:12:03 pm

I have found that 2 + 2i and therefore 2-2i are solutions of this, but how do I find the remaining two? I've tried long division but it's not working for me..

z^4 - 6z^3 + 19z^2 - 28z +24 = 0

Title: Re: Complex Numbers
Post by: max payne on May 03, 2012, 09:35:46 pm

multiply the two factors together to get a quadratic and then try long division:
eg. (z-(2+2i))*(z-(2-2i))= (z^2-4z+8)
I think you can take it from here...

Title: Re: Complex Numbers
Post by: Becky2012 on May 04, 2012, 05:00:45 pm

Yeah I figured it out today, turns out I had expanded the two factors wrong making my division a lot harder than it should have been haha.. Thanks though!