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Math Help - complex number help

  1. #1
    MHF Contributor
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    complex number help

    The complex number z=\frac{1+i}{3-2i}.Write down the argument for z^2 and find the exact value of |z^2| .

    My working :

    z=\frac{1}{13}+\frac{5}{13}i
    z^2=(\frac{1}{13}+\frac{5}{13}i)^2
    z^2=\frac{-24+10i}{169}

    Up to here , did i make any mistake . My final answer is wrong so i am wondering where my mistake is . Can someone pls check for me ? Thanks .
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  2. #2
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    Hello, mathaddict!

    Given: . z\:=\:\frac{1+i}{3-2i}.

    Find the argument for z^2 and find the exact value of |z^2| .


    My working:

    z\:=\:\frac{1}{13}+\frac{5}{13}i

    z^2\:=\:\left(\frac{1}{13}+\frac{5}{13}i\right)^2 \;=\;\frac{-24+10i}{169}


    Up to here, did i make any mistakes? .
    . . . no

    My final answer is wrong. .
    You could have shown it to us.

    We have: . z^2 \;=\;-\frac{24}{169} + \frac{10}{169}i


    The argument \theta is given by: . \tan\theta \;=\;\frac{\frac{10}{169}}{-\frac{24}{169}} \;=\;-\frac{5}{12}

    Hence: . \theta \;=\;\tan^{-1}\left(\text{-}\tfrac{5}{12}\right) \;\approx\; 2.7468\text{ radians} \;\approx\;157.38^o


    The magnitude of z^2 is: . |z^2| \;=\;\sqrt{\left(\frac{\text{-}24}{169}\right)^2 + \left(\frac{10}{169}\right)^2}\;= \;\sqrt{\frac{676}{169^2}}

    Therefore: . |z^2| \;=\;\frac{26}{169} \;=\;\frac{2}{13}

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