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Proving that a complex number is the root of an equation. and finding the other roots Watch

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    Prove that 1+i is a root of the equation z(^4)+3z(^2)-6z+10=0

    So I took z=1+i and proved that LHS=0, so it's proved that 1+i is a root. But it next asks me to find the other roots of the equation.

    I usually find the roots by substituting values of z randomly into the calculator for the equation and if say z=1 gives the equation as 0, I write down (z-1) is a factor.

    Question 1. Is there any other better way to do this?

    Question 2. For this particular equation no real numbers seem to work to give LHS=0, do I have to use 1+i in the equation to find the other roots? if so how?

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    apologies for the seemingly simple questions, i'm self studying
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    1-i is another root because it's the complex conjugate.
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    (Original post by gunmetalpanda)
    Prove that 1+i is a root of the equation z(^4)+3z(^2)-6z+10=0

    So I took z=1+i and proved that LHS=0, so it's proved that 1+i is a root. But it next asks me to find the other roots of the equation.

    I usually find the roots by substituting values of z randomly into the calculator for the equation and if say z=1 gives the equation as 0, I write down (z-1) is a factor.

    Question 1. Is there any other better way to do this?

    Question 2. For this particular equation no real numbers seem to work to give LHS=0, do I have to use 1+i in the equation to find the other roots? if so how?

    Spoiler:
    Show
    apologies for the seemingly simple questions, i'm self studying
    In the case of complex numbers, if a number is a root of the equation, then it's conjugate will also be a root of the equation. Are you aware of what a complex conjugate is? Use these roots to form factors and thus factorise the expression (it will help if you multiplied the each conjugate factor of the pair together to get the polynomial as a product of two quadratics.)

    EDIT: Provided that we have real coefficients. Thanks Daniel.
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    (Original post by Farhan.Hanif93)
    In the case of complex numbers, if a number is a root of the equation, then it's conjugate will also be a root of the equation.
    (Provided the coefficients are real)
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    (Original post by Daniel Freedman)
    (Provided the coefficients are real)
    God, I should have really said that... I definitely had that condition in my mind when posted, honest! :p:
 
 
 
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