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Math Help - roots of complex numbers

  1. #1
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    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
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  2. #2
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    Quote Originally Posted by philyc86 View Post
    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?
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  3. #3
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    Quote Originally Posted by philyc86 View Post
    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.
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  4. #4
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    Quote Originally Posted by shawsend View Post
    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.
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  5. #5
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    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.
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  6. #6
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    Quote Originally Posted by mr fantastic View Post
    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.
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  7. #7
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    Quote Originally Posted by shawsend View Post
    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!
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  8. #8
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    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 ?
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  9. #9
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    Quote Originally Posted by philyc86 View Post
    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.
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