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
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.