simple complex quotient! wtf! need help!

Discussion in 'Physics & Math' started by camilus, Sep 24, 2008.

  1. camilus the villain with x-ray glasses Registered Senior Member

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    895
    sun, I dont know why Im stuggling with a simple freakin problem. Its not even for me its for my friend who's in college algebra, but for some reaon I cant get the correct answer.

    \({9 - 11i \over 6i}\)

    I multiplied by the conjugate twice and got -11/6 + 3i/2

    what did I do wrong?
     
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  3. mathman Valued Senior Member

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    Multiply the numerator and denominator by -i. The result will be -11/6 - 3i/2.
     
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  5. Sciencelovah Registered Senior Member

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    i x i = -1

    so:

    \({9 - 11i \over 6i}\) x \(i \over i\)

    = \({9i + 11 \over -6}\)

    = \({-9i \over 6}\) - \(11 \over 6\)

    = \({-3i \over 2}\) - \(11 \over 6\)
     
    Last edited: Sep 24, 2008
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  7. Sciencelovah Registered Senior Member

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    Wait I make mistake on the 2nd row from bottom. Now is correct. Mathman was correct

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  8. rpenner Fully Wired Valued Senior Member

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    Review:
    Multiplication: (a + bi)(c + di) = (ab-bd)+(ad+bc)i
    The rule to normalize a ratio of complex numbers is to multiply top and bottom by the conjugate of the denominator.
    Then everything is mechanical but you have to carefully track the minus signs.

    So \(\begin{eqnarray} {{9-11i}\over{6i}} & = & {{9-11i}\over{6i}}\times{{-6i}\over{-6i}} \\ & = & {{(9 + (-11)i)(0 + (-6)i)}\over{(0+6i)(0+(-6)i)}} \\ & = & {{(9 \times 0 - (-11)\times(-6)) + (9 \times (-6) + 0 \times (-11))i}\over{(0 \times 0 - (6)\times(-6)) + (0 \times (-6) + 6 \times 0)i}} \\ & = & {{-66 + (-54)i}\over{36}} \\ & = & {{-11 + (-9)i}\over{6}} \\ & = & -\frac{11}{6}-\frac{3}{2}i \end{eqnarray}\)
     
  9. camilus the villain with x-ray glasses Registered Senior Member

    Messages:
    895
    thanks fellas i got a minus sign wrong, I ha 3i/2 instead of -3i/2.
     

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