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Homework Help: Find Re(z)

  1. Feb 17, 2015 #1
    1. The problem statement, all variables and given/known data
    Find the real part of z, given that 1/z = 1/R + 1/(iωL)

    2. Relevant equations

    3. The attempt at a solution
    I rearranged to find z, and got that z = R + iωL. So that would make the Re(z)=R, wouldn't it?? It's wrong, but I don't have a clue why!
  2. jcsd
  3. Feb 17, 2015 #2


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    When you rearranged, you got the wrong result. Can you show working for that please?
    (You can't just turn all the fractions upside down.)
    Last edited: Feb 17, 2015
  4. Feb 17, 2015 #3
    Ohhhhh soooo stupid... forgot to multiply the LHS when finding a common denominator.
  5. Feb 17, 2015 #4
    Actually, hang on... here's the rearrangement.

    1/z = 1/R + 1/(iwL) Multiply by R:

    R/z = 1 + R/(iwL) Multiply by iwL:

    RiwL/z = iwL + R

    So z = (RiwL) / (iwL + R)?? There's clearly something wrong but I can't see what... Do I need to realise the denominator?
  6. Feb 17, 2015 #5
    That's true, but you're asked to find Re(z). You can use this useful identity that comes up a lot in electrical engineering:
    z = \frac{z \overline{z}}{\overline{z}} = \frac{|z|^2}{\overline{z}}
    Along with:
    \mathrm{Re}\!\left(\frac{1}{\overline{z}}\right) = \mathrm{Re}\!\left(\frac{1}{z}\right)
    Last edited: Feb 17, 2015
  7. Feb 17, 2015 #6


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    Yep. Fine so far. You're quite right - what is wrong is that you still can't separate the Re and Im parts.

    Miles has put something above (which I don't understand) and if that helps you, great.

    For me, the next step is just plain maths, nothing to do with electronics. If you want to separate this expression into Re & Im, you need to have a real denominator and the standard maths 'trick' is to multiply a complex denominator by its complex conjugate.

    Every complex number has a conjugate and when you multiply it by the conjugate you get a real number - the i's disappear.
    Any fraction can be multiplied by 1 (or x/x since anything divided by itself is 1), so you multiply your fraction by conjugate/conjugate
    Now the conjugate is dead simple: conjugate of A + iB is A - iB or conjugate of C - iD is C + iD (negate the Im part)
    So you need to multiply your z expression by (R - iωL ) / ( R - iωL)

    And just to complete the maths, remember the difference of two squares,
    x2 - y2 = (x-y)(x+y)
    So (A + iB)(A - iB) = A2 - (i)2B2
    = A2 - (-1)B2
    = A2 + B2
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