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Math Help - complex number question

  1. #1
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    complex number question

    Find the square root of the following complex number :

    6+8i

    \sqrt{6+8i}=a+bi

    6+8i=a^2-b^2+2abi

     <br />
a^2-b^2=6 ----1<br />
    2ab=8 ---- 2

    From 2 , b=\frac{4}{a}

    Substitute into 1 : a^2-\frac{16}{a^2}=6 which gives me
    a=\pm2\sqrt{2}

    then b=\pm\frac{2}{\sqrt{2}}

    did i get the correct answers for a and b ? THanks for helping .
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  2. #2
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    Quote Originally Posted by thereddevils View Post
    Find the square root of the following complex number :

    6+8i

    \sqrt{6+8i}=a+bi

    6+8i=a^2-b^2+2abi

     <br />
a^2-b^2=6 ----1<br />
    2ab=8 ---- 2

    From 2 , b=\frac{4}{a}

    Substitute into 1 : a^2-\frac{16}{a^2}=6 which gives me
    a=\pm2\sqrt{2}

    then b=\pm\frac{2}{\sqrt{2}}

    did i get the correct answers for a and b ? THanks for helping .
    Yup, looks good.
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  3. #3
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    Quote Originally Posted by thereddevils View Post
    Find the square root of the following complex number :

    6+8i

    \sqrt{6+8i}=a+bi

    6+8i=a^2-b^2+2abi

     <br />
a^2-b^2=6 ----1<br />
    2ab=8 ---- 2

    From 2 , b=\frac{4}{a}

    Substitute into 1 : a^2-\frac{16}{a^2}=6 which gives me
    a=\pm2\sqrt{2}

    then b=\pm\frac{2}{\sqrt{2}}

    did i get the correct answers for a and b ? THanks for helping .
    I'm trying to solve a similar problem. Could you please show me how did you get 'a'??
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  4. #4
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    Quote Originally Posted by Math's-only-a-game View Post
    I'm trying to solve a similar problem. Could you please show me how did you get 'a'??
    Multiply each side of the equation by a^2 :
    (a^2)^2-6a^2-16=0

    Let x=a^2, giving x^2-6x-16=0
    and solve the quadratic. But be aware that x=a^2\geq 0. So you ought to keep only the positive root.
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