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Math Help - dealing with -i

  1. #1
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    dealing with -i

    I know its not acceptable to have i in the denominator of an equation, however I'm not sure what to do when I have -i, do I need to get rid of the - (negative sign) sign?

    My example is that using the quadratic I've gotten to this point z= (2 i sqrt(3))/2i. I know I need to get rid of the i in the denominator so I multiply the whole thing by i/i then I get -i + 1/2 (i*sqrt(3))/2. I'm wondering if I need to get rid of the negative sign in front of the i or if I'm doing something wrong, or can i just leave it as is.

    Thanks
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by GalenWilson View Post
    I know its not acceptable to have i in the denominator of an equation, however I'm not sure what to do when I have -i, do I need to get rid of the - (negative sign) sign?

    My example is that using the quadratic I've gotten to this point z= (2 i sqrt(3))/2i. I know I need to get rid of the i in the denominator so I multiply the whole thing by i/i then I get -i + 1/2 (i*sqrt(3))/2. I'm wondering if I need to get rid of the negative sign in front of the i or if I'm doing something wrong, or can i just leave it as is.

    Thanks
    I cant really te,ll what equation that is but if it is helpful -i=i\cdot{-1}=i\cdot{i^2}=i^3
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  3. #3
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    Rephrase

    The questions starts out as iz^2 + (2 + i)z + 1. Using the quadratic equation it becomes z= (-2 - i sqrt(3))/2i. I know I need to get rid of the i in the denominator by multiplying the whole problem by i/i. I *believe* this leaves me with -i + 1/2 (i*sqrt(3))/2.

    My question is A) am I wrong and B) if not then what do I do (if anyhting with -i?

    I think I'm just confused.
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by GalenWilson View Post
    The questions starts out as iz^2 + (2 + i)z + 1. Using the quadratic equation it becomes z= (-2 - i sqrt(3))/2i. I know I need to get rid of the i in the denominator by multiplying the whole problem by i/i. I *believe* this leaves me with -i + 1/2 (i*sqrt(3))/2.

    My question is A) am I wrong and B) if not then what do I do (if anyhting with -i?

    I think I'm just confused.
    \frac{-2 - i \pm \sqrt{3}}{2i}

    = \frac{-2 - i \pm \sqrt{3}}{2i} \cdot \frac{i}{i}

    = \frac{-2i - i^2 \pm i\sqrt{3}}{2i^2}

    = \frac{-2i +1 \pm i\sqrt{3}}{-2}

    = i - \frac{1}{2} \pm \frac{i\sqrt{3}}{2}

    Do you need to get rid of the - when it's in the denominator? (ie. Must you do that last step?) No, but I'd say it's likely a preferred form.

    -Dan
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  5. #5
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    In general here is are very very useful facts about complex numbers:
    \frac{1}{z} = \frac{{\overline z }}{{\left| z \right|^2 }}, so \frac{1}{{a + bi}} = \frac{{a - bi}}{{a^2  + b^2 }}.


    Examples: \frac{1}{{ - 2 + 3i}} = \frac{{ - 2 - 3i}}{{13}}
    \left( {4 - 5i} \right)^{ - 4}  = \frac{1}{{\left( {4 - 5i} \right)^4 }} = \left[ {\frac{{4 + 5i}}{{41}}} \right]^4
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