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November 01, 2025, 08:04:22 pm
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Author Topic: Complex Numbers  (Read 734 times)  Share 

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Becky2012

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Complex Numbers
« on: May 03, 2012, 09:12:03 pm »
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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
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max payne

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Re: Complex Numbers
« Reply #1 on: May 03, 2012, 09:35:46 pm »
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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...
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Becky2012

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Re: Complex Numbers
« Reply #2 on: May 04, 2012, 05:00:45 pm »
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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!
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