hi there

how would i find all thew roots of a polynomial like this

z^4+z^3+(1-j)z+(j-1)=0

i have looked all over but cant find another example that includes anything like the (1-j)z term

thankyou

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- Nov 7th 2008, 07:34 AMphilyc86roots of complex numbers
hi there

how would i find all thew roots of a polynomial like this

z^4+z^3+(1-j)z+(j-1)=0

i have looked all over but cant find another example that includes anything like the (1-j)z term

thankyou - Nov 7th 2008, 01:01 PMmr fantastic
- Nov 7th 2008, 02:31 PMshawsend
How about doing it the old-school way:

(1) first convert it to the reduced quartic:

$\displaystyle y^4+qy^2+ry+s=0$ via a change of variable $\displaystyle z=y-1/4 $. Doing this I get:

$\displaystyle y^4-3/8y^2+(1/8+d)y+f=0;\quad d=1-i,\quad f=i-1-\frac{1-i}{4}-\frac{3}{256}$

Substituting this into the resolvent cubic:

$\displaystyle 64k^6+32qk^4+4(q^2-4s)k^2-r^2=0$

yields:

$\displaystyle 64k^6-12k^4+(9/16-16f)k^2-(1/64+d/4+d^2)=0$

This can now be solved as a cubic using the old-school approach to cubics. Make all the back substitutions to arrive at the roots. Lotta' work. Maybe an easier way though. - Nov 7th 2008, 02:34 PMmr fantastic
- Nov 7th 2008, 02:38 PMshawsend
I did solve it in Mathematica. The symbolic (exact) roots are very messy, too long for a single one to fit in the space allowed by Latex here.

- Nov 7th 2008, 02:40 PMshawsend
- Nov 7th 2008, 02:50 PMmr fantastic
- Nov 9th 2008, 02:27 AMphilyc86
hi

thanks for your replies .I did make a typo ,but only with the powers of the z terms ,which are 5 then 4 ,rather than 4 and 3 .any more ideas ? - Nov 9th 2008, 02:46 AMmr fantastic