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Complex number exponential subtraction

  1. Feb 1, 2005 #1
    Hi all,

    Im having a bit of trouble with a question. I have to convert:

    [tex]Ke^{j\delta} - Ke^{j\psi}[/tex]

    Into the form


    This is the second part of the question, the first part was an addition instead of subtraction which i managed by using this formula:

    [tex]z_1 + z_2 = K(e^{j\delta} + e^{j\psi}) = Ke^{j(\delta + \psi)/2}(e^{j(\delta - \psi)/2} + e^{-j(\delta - \psi)/2}) = 2K\cos((\delta - \psi)/2).e^{j(\delta + \psi)/2}[/tex]

    I cant really see where to go with the subtraction though...is it maybe to do with a sin rule? To be honest i dont fully understand the formula above but it was given to us...Ive fiddled around with the maths for a while but its totally headbanging :( Hoping someone can help!
  2. jcsd
  3. Feb 1, 2005 #2


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    Since you already have a formula for sums, how about using the fact that [itex]e^{i(\psi+ \pi)}= e^{i\psi}e^{i\pi}= -e^{i\psi}[/itex]? In other words, replace [itex]\psi[/itex] with [itex]\psi+ \pi[/itex] in your formula.
  4. Feb 1, 2005 #3


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    Homework Helper

    Okay.It's not difficult.Jst follow the same pattern using the half angle

    [tex] D=z_{1}-z_{2}=Ke^{j(\frac{\delta+\psi}{2})}[e^{j(\frac{\delta-\psi}{2})}-e^{j(\frac{\psi-\delta}{2})}]=Ke^{j(\frac{\delta+\psi}{2})}[e^{j(\frac{\delta-\psi}{2})}-e^{-j(\frac{\delta-\psi}{2})}]=[2jK\sin(\frac{\delta-\psi}{2})]e^{j(\frac{\delta+\psi}{2})} [/tex]


  5. Feb 1, 2005 #4
    brilliant thanks for your help! :)
  6. Apr 25, 2006 #5
    URGENT: Can anyone help me with this?

    Can anyone tell me what difference it makes to the subtraction/addition formulae when the base factors are not the same? i.e


    I have fiddled with the above formulae, but cant seem to get anything sensible looking from it... And I need to solve this quite urgently!!!

    Thanks in advance...
    Last edited: Apr 25, 2006
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