1. Limited time only! Sign up for a free 30min personal 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!

Proof of multiplicative inverse of complex number

  1. Sep 2, 2009 #1
    1. The problem statement, all variables and given/known data
    I'm trying to derive the formula for the multiplicative inverse of a complex number. I say that:

    z=a+bi
    z-1=u+vi

    and zz-1 must be 1 so

    zz-1=(a+bi)(u+vi)
    =(au-bv)+(av+bu)i
    =1

    So in order for zz-1=1, (au-bv) must be 1 and (av+bu) must be 0. My teacher has solved this as:

    z-1= a/(a2+b2) - bi/(a2+b2)

    I just don't know how to get that from au-bv=1 and av+bu=0

    2. The attempt at a solution
    Don't know how to get the last part... major brain fog. Please help?
     
  2. jcsd
  3. Sep 2, 2009 #2
    It's just a sytem of two linear equations in the two unknowns u and v. You can set it up as a matrix and do row reduction or you can do it the old fashioned way.
     
  4. Sep 2, 2009 #3

    Mark44

    Staff: Mentor

    I don't see that you have used the fact that z-1 = 1/(a + bi). Multiply the latter expression by 1 in the form of (a - bi)/(a - bi). See if that helps.
     
  5. Sep 2, 2009 #4
    Thanks Mark, I got the answer using your method :)

    But, if I were to solve this using linear algebra, what would my matrix look like? I think:

    a -b 1
    b a 0

    That doesn't look right, and I'm having difficulties solving this. Any pointers? Thanks in advance.
     
    Last edited: Sep 2, 2009
  6. Sep 2, 2009 #5
    a matrix is overkill for this one.

    you have au - bv = 1 and av + bu = 0

    so u = (1 + bv)/a. substitute this into the second one to get


    [tex]av + b\frac{(1 + bv)}{a} = 0[/tex]

    [tex]v(a + \frac{b^2}{a}) = -\frac{b}{a}[/tex]

    [tex]v\frac{a^2 + b^2}{a} = -\frac{b}{a}[/tex]


    you can probably take it from there.

    If you do want to do it with matrices, the first step is to multiply the first row by -b/a and add it to the second row. That should get you going.
     
    Last edited: Sep 2, 2009
  7. Sep 2, 2009 #6
    You're a champ, thanks a lot!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Proof of multiplicative inverse of complex number
  1. Proof complex numbers (Replies: 8)

Loading...