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    Hi

    Solve the equation: z^2 + 2z + 10 = 0
    Find the modulus and argument of each root.

    :confused:
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    (Original post by SeekerOfKnowledge)
    Hi

    Solve the equation: z^2 + 2z + 10 = 0
    Find the modulus and argument of each root.

    :confused:
    first use the quadratic equation/formulae

    z = \frac{-b \pm \sqrt{b^2-4ac}}{2a}

    z = \frac{-2 \pm \sqrt{4-4(1)(10)}}{2(1)}

    z = \frac{-2 \pm \sqrt{-36}}{2}

    z = \frac{-2 \pm 6i}{2}

    z=-1 \pm3i

    use pythagoras to find the modulus,

    so

    \sqrt {(-1)^{2} + 3^{2}}= \sqrt 10

    and sohcahtoa to find the argument.

    \pi - [tan^{-1}(\frac{3}{-1})] (pi minus answer since it's in the 2nd quadrant.)

    74.7066....
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    Use the quadratic formula to get z in Cartesian form.
    Do you know how to get modulus and argument from there?
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    just solve it like you would any quadratic equation.. and you'll find that b^2 - 4ac < 0...hence giving you a complex number

    Spoiler:
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    Root-36 = iRoot36 = 6i
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    (Original post by HerrWarum)
    first use the quadratic equation/formulae

    z = \frac{-b \pm \sqrt{b^2-4ac}}{2a}

    z = \frac{-2 \pm \sqrt{4-4(1)(10)}}{2(1)}

    z = \frac{-2 \pm \sqrt{-36}}{2}

    z = \frac{-2 \pm 6i}{2}

    z=-1 \pm3i

    use pythagoras to find the modulus,

    so

    \sqrt {(-1)^{2} + 3^{2}}= \sqrt 10

    and sohcahtoa to find the argument.

    \pi - [tan^{-1}(\frac{3}{-1})] (pi minus answer since it's in the 2nd quadrant.)

    74.7066....

    aha i see, thanks

    yes Slumpy i think so:

    arctan = (pi) - 3/1

    modulus = sqrt 1^2 + 3^2

    right?




    erm Ewan im a little confussed with what you have done
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    (Original post by SeekerOfKnowledge)
    aha i see, thanks

    yes Slumpy i think so:

    arctan = (pi) - 3/1

    modulus = sqrt 1^2 + 3^2

    right?
    Modulus is right.
    You've slightly missed what was being done on the argument, what was meant was:
    argz= pi - arctan(3/1)
    Although this misses the other root, which has argument:
    argz=pi+ arctan(3/1(

    (I've ignored the minus sign as I've considered that by taking the correct quadrant)
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    (Original post by SeekerOfKnowledge)
    erm Ewan im a little confussed with what you have done
    http://scholar.hw.ac.uk/site/maths/topic11.html work through that site, its useful and has diagrams explaining what you do in different quadrants..

    I was just explaining at Root-36 is 6i & that its only complex because b^2 - 4ac < 0... read what Herr wrote

    Just remember that the arguement is always measured from the positive x axis to the line. Its basically just a convention, since there are infinitely many ways of writing the same arguement.
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    (Original post by Ewan)
    http://scholar.hw.ac.uk/site/maths/topic11.html work through that site, its useful and has diagrams explaining what you do in different quadrants..

    I was just explaining at Root-36 is 6i & that its only complex because b^2 - 4ac < 0... read what Herr wrote

    Just remember that the arguement is always measured from the positive x axis to the line. Its basically just a convention, since there are infinitely many ways of writing the same arguement.
    oh i see. Thank you for the site.



    Slumpy
    oh yeah forgot to add the arctan,
    but would that give me two arguments?
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    (Original post by Ewan)
    http://scholar.hw.ac.uk/site/maths/topic11.html work through that site, its useful and has diagrams explaining what you do in different quadrants..

    I was just explaining at Root-36 is 6i & that its only complex because b^2 - 4ac < 0... read what Herr wrote

    Just remember that the arguement is always measured from the positive x axis to the line. Its basically just a convention, since there are infinitely many ways of writing the same arguement.
    thanks for you help people
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    Did u quote me then delete your post? :p:
 
 
 
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