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Calculating Re(z)

  1. Feb 17, 2015 #1
    In the equation 1/z = 1/R +1/(iwL), why isn't the real part just R?

    Unless my rearrangement is wrong, z=R+iwl, so Re(z) should definitely be R!
     
  2. jcsd
  3. Feb 17, 2015 #2

    HallsofIvy

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    Yes, your rearrangement is wrong. [itex]\frac{1}{a}= \frac{1}{b}+ \frac{1}{c}[/itex] does NOT give "a= b+ c". As a check, suppose b= c= 2. The 1/a= 1/2+ 1/2= 1 so a= 1 But b+ c= 4, not 1.

    From [tex]\frac{1}{z}= \frac{1}{R}+ \frac{1}{iWL}[/tex], multiply on both sides by all denominators, RiWLz, to get RiWL= iWLz+ Rz= (R+ iWL)z. Solve that for z to find the real and imaginary part.
     
  4. Feb 17, 2015 #3

    wabbit

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    So you're saying 1/(R+iwL)=1/R+1/(iwL) ?
     
  5. Feb 17, 2015 #4
    Yes. I did actually think that was weird, not sure why I didn't check it. Let me try that rearrangement again...
     
  6. Feb 17, 2015 #5
    z = -R2ω2L2/(iωL+R) is my second attempt!


    Edit: maybe not. L squared too. And that makes life more difficult! What do I do about i on the denominator? Multiply by the conjugate?
     
    Last edited: Feb 17, 2015
  7. Feb 17, 2015 #6

    HallsofIvy

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    Pretty standard: multiply both numerator and denominator by the complex conjugate of the denominator.
    [tex]\frac{a+ bi}{c+ di}= \frac{a+ bi}{c+ di}\frac{c- di}{c- di}[/tex][tex]= \frac{ac+ bd+ (bc- ad)i}{c^2+ d^2}=[/tex][tex]\frac{ac+ bd}{c^2+ d^2}+ \frac{bc- ad}{c^2+ d^2} i[/tex]

    Here you want to multiply numerator and denominator by [tex]R- i\omega L[/tex].
     
  8. Feb 17, 2015 #7
    z = RiwL/(R+iwL)
    Realise the denominator:
    (iR2wL+Rw2L2)/(R2+w2L2)
     
  9. Feb 17, 2015 #8

    wabbit

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    You can do better than that (your first line is OK though).
    Try this as a warm up : forget the exercise, just compute [tex]\frac{1}{a+ib}[/tex]. Then go back to z.
     
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