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Homework Help: 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:


    and zz-1 must be 1 so


    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


    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!
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