# roots of complex numbers

• Nov 7th 2008, 07:34 AM
philyc86
roots 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 PM
mr fantastic
Quote:

Originally Posted by philyc86
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

Is there a typo? Could it be z^4 + z^3 + (1 - j)z + (1 - j) = 0?
• Nov 7th 2008, 02:31 PM
shawsend
Quote:

Originally Posted by philyc86
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:

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

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

Substituting this into the resolvent cubic:

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

yields:

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

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 PM
mr fantastic
Quote:

Originally Posted by shawsend
How about doing it the old-school way:

(1) first convert it to the reduced quartic:

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

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

Substituting this into the resolvent cubic:

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

yields:

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

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.

I thought of this and then figured there was a much better chance of a typo than of this solution pathway being intended. Hopefully the OP will clarify.
• Nov 7th 2008, 02:38 PM
shawsend
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 PM
shawsend
Quote:

Originally Posted by mr fantastic
Is there a typo? Could it be z^4 + z^3 + (1 - j)z + (1 - j) = 0?

Ok. I see what you mean: the root are much simpler that way.
• Nov 7th 2008, 02:50 PM
mr fantastic
Quote:

Originally Posted by shawsend
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.

Aha! So the chance gets better and better by the minute lol!
• Nov 9th 2008, 02:27 AM
philyc86
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 AM
mr fantastic
Quote:

Originally Posted by philyc86
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 ?

Yes. There's most likely a typo (like the one I've flagged) in the original question. Speak to your instructor.