1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    Merlin3189

    User Avatar
    Gold Member

    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

    Merlin3189

    User Avatar
    Gold Member

    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
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Find Re(z)
  1. Find all z (Replies: 6)

  2. Finding z-scores (Replies: 5)

Loading...