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 November 5th, 2008, 07:05 AM #1 Newbie   Joined: Oct 2008 Posts: 2 Thanks: 0 Finding roots of a complex numbers equation I cant seem to fully figure a problem like this one out. How is it done? z^5-z^4+(1-j)z+(j-1)=0 Thanks
November 5th, 2008, 02:19 PM   #2
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Re: Finding roots of a complex numbers equation

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 Originally Posted by nisko I cant seem to fully figure a problem like this one out. How is it done? z^5-z^4+(1-j)z+(j-1)=0 Thanks
z-1 is a factor. Your equation is then (z-1)(z^4 +1 -j)=0. You should be able to handle this.

 November 6th, 2008, 09:52 AM #3 Newbie   Joined: Oct 2008 Posts: 10 Thanks: 0 Re: Finding roots of a complex numbers equation z^5-z^4+(1-j)z+(j-1)=0 goes in this way: z^4 (z-1) + (1-j)z - (1-j) = 0 <=> z^4 (z-1) + (1-j) (z-1) = 0 <=> and then factorize (z-1) hence the (z-1) (z^4+1-j) = 0 then z= 1 OR z^4=-1+j (...) now it is pretty easy. (z^4 must have only 4 roots. put it in polar form and it's done!)

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