complex no. help

Discussion in 'Homework Help' started by ecjohnny, Nov 29, 2008.

  1. ecjohnny

    Thread Starter Senior Member

    Jul 16, 2005
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  2. leftyretro

    Active Member

    Nov 25, 2008
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  3. blazedaces

    Active Member

    Jul 24, 2008
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    What step specifically are you having an issue with?

    -blazed
     
  4. Papabravo

    Expert

    Feb 24, 2006
    10,140
    1,789
    If you multiply both the numerator and the denominator by the complex conjugate of the denominator, then the denominator will beome a real number since j*(-j) = -(j^2) = 1
     
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