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Complex numbers help

just making notes on finding roots of quartics etc with complex numbers, are these statements correct:?
1. (z-(x))(z-(y)) - where x and y are roots of the equation , expanding this will give you the original equation, 3 brackets for cubic and 4 for quartic.
2. (z-x) ( y ) - X is a known root of the equation, where Y is an unknown factor of the equation, you calculate Y based on the original equation and the given root, and then if Y is a quadratic you solve for 2 roots.
so if i know the roots but not the equation, i use method 1 , if i know the equation but not the roots i use method 2?
are these statements entirely correct as i dont want to write it down if its a bit wrong
(edited 5 years ago)
Reply 1
Original post by Gent2324
just making notes on finding roots of quartics etc with complex numbers, are these statements correct:?
1. (z-(x))(z-(y)) - where x and y are roots of the equation , expanding this will give you the original equation, 3 brackets for cubic and 4 for quartic.
2. (z-x) ( y ) - X is a known root of the equation, where Y is an unknown factor of the equation, you calculate Y based on the original equation and the given root, and then if Y is a quadratic you solve for 2 roots.
so if i know the roots but not the equation, i use method 1 , if i know the equation but not the roots i use method 2?
are these statements entirely correct as i dont want to write it down if its a bit wrong


1. Yes, but you won't have the z^3 or z^4 coefficient. For the cubic you have 3 roots but 4 coeffs. Assuming the coeff of z^3 is 1, then you're ok. Similar for quartic.

Don't fully understand your explanation in 2, but sounds about right, with a similar restriction to 1. Can you give an example?
(edited 5 years ago)
Original post by mqb2766
1. Yes, but you won't have the z^3 or z^4 coefficient. For the cubic you have 3 roots but 4 coeffs. Assuming the coeff of z^3 is 1, then you're ok. Similar for quartic.

Don't fully understand your explanation in 2, but sounds about right. Can you give an example?


z^3 - 6z^2 +21z - 26
given 2 is a root, solve f(z) completely.

(z-2) ( )
coefficent of z^3 is 1, so must be z^2 as the first term in the bracket, the number on its own is -26, so -2 multiplied by a number (13) is -26
(z-2)(z^2 +13)
we are left to find the z^2's which is -6, we know that -2 mulitplied by z^2 is -2z^2 , so we just need another -4z^2 to get to -6z^2 , which means z multiplied by -4z is -4z^2, so we are left with:
(z-2)(z^2 - 4z + 13)
solve the quadratic to give 2 roots, which gives 3 overall.
i havent tried this with quartics yet, but i think it just means you have z^2 in the first bracket
Reply 3
Original post by Gent2324
z^3 - 6z^2 +21z - 26
given 2 is a root, solve f(z) completely.

(z-2) ( )
coefficent of z^3 is 1, so must be z^2 as the first term in the bracket, the number on its own is -26, so -2 multiplied by a number (13) is -26
(z-2)(z^2 +13)
we are left to find the z^2's which is -6, we know that -2 mulitplied by z^2 is -2z^2 , so we just need another -4z^2 to get to -6z^2 , which means z multiplied by -4z is -4z^2, so we are left with:
(z-2)(z^2 - 4z + 13)
solve the quadratic to give 2 roots, which gives 3 overall.
i havent tried this with quartics yet, but i think it just means you have z^2 in the first bracket


Yes, you divide by the known factor and find the roots of the remaining quotient. You still have a constant multiplier in the original equation that you don't know, but assuming the coeff of z^3 is 1 (or the coeff of z^4 is 1 for a quartic) keeps things happy.
(edited 5 years ago)

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