# Thread: roots of complex numbers

1. ## 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

2. 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?

3. 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:

$\displaystyle y^4+qy^2+ry+s=0$ 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.

4. Originally Posted by shawsend
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.
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.

5. 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.

6. 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.

7. 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!

8. 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 ?

9. 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.