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Homework Help: Problem of math Proof

  1. May 8, 2010 #1
    1. The problem statement, all variables and given/known data
    How do I show that if (a+ib)2 = (c+id)2 then (a+ib) = [tex]\pm[/tex](c+id)

    2. Relevant equations



    3. The attempt at a solution
    (a+ib)(a+ib) = a2 + 2aib -b2
    (c+id)(c+id) = c2 + 2cid - d2
    Stuck after this.
     
  2. jcsd
  3. May 8, 2010 #2

    gabbagabbahey

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    Re: Proof

    Well, if a,b,c and d are all real numbers, then you must have a2-b2=c2-d2 and ab=2cd (for two complex number to be equal, their real parts must be equal and their imaginary parts must be equal)
     
  4. May 8, 2010 #3
    Re: Proof

    Don't you mean ab = cd. So b = cd/a then (cd/a)2 - b2 = c2 - d2
    a4 - c2d2 = a2c2 - a2 - d2
    (a2 - c2)(a2 + d2) = 0
    Am I on the right track
     
  5. May 8, 2010 #4

    gabbagabbahey

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    Re: Proof

    Yes. Now realize that for (a2 - c2)(a2 + d2) to be zero, either a2= c2 or c2=-d2....but "a", "c" and "d" are real numbers, so a=___?
     
  6. May 8, 2010 #5

    jgens

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    Re: Proof

    If [itex](a+bi)^2=(c+di)^2[/itex], then clearly [itex](a+bi)^2 - (c+di)^2 = 0[/itex]. Factor this expression and use the zero product property to arrive at the desired result.
     
  7. May 8, 2010 #6

    gabbagabbahey

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    Re: Proof

    That works fine, provided you've already proven that

    [tex]z_1z_2=0\implies z_1=0\;\;\;\text{or}\;\;\;z_2=0[/itex]

    for complex numbers.
     
  8. May 9, 2010 #7
    Re: Proof

    a2 = -d2 & a2 = c2
    a = [tex]\pm[/tex]id & a = [tex]\pm[/tex]c
     
    Last edited: May 9, 2010
  9. May 9, 2010 #8

    gabbagabbahey

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    Re: Proof

    Right, but you can immediately throw away the solution [itex]a=\pm id[/itex] since both [itex]a[/itex]and [itex]d[/itex] are supposed to be real numbers....What does the other solution give you for [itex]b[/itex] when you plug it back into [itex]b=\frac{cd}{a}[/itex]?
     
  10. May 9, 2010 #9
    Re: Proof

    b = [tex]\pm[/tex]d
    What do we plug that into?
    Since we know what a & b is hence (a+ib) = ([tex]\pm[/tex]c + i[tex]\pm[/tex]d) then factorise to get (a+ib) = [tex]\pm[/tex](c+id)
     
    Last edited: May 9, 2010
  11. May 9, 2010 #10

    gabbagabbahey

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    Re: Proof

    Yeah, basically. If [itex]a=\pm c[/itex] and [itex]b=\pm d[/itex], then [itex](a+ ib)=\pm(c + i d)[/itex] and you're done.:smile:
     
  12. May 9, 2010 #11
    Re: Proof

    Awesome.
    Cheers.
    What if I wanted to do it in polar form.
     
  13. May 10, 2010 #12

    gabbagabbahey

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    Re: Proof

    In polar form, you would basically want to show that [itex](r_1e^{i\phi_1})^2=(r_2e^{i\phi_2})^2[/itex] implies [itex]r_1e^{i\phi_1}=\pm r_2e^{i\phi_2}[/itex] and for that, you would use the fact that [itex]e^{i\phi_2}=e^{i(\phi_2+2\pi n}[/itex] to finds the two unique roots of the equation.
     
  14. May 10, 2010 #13
    Re: Proof

    What if I wanted to go (a+ib)2 = r12(cos(2[tex]\vartheta[/tex]) + isin(2[tex]\vartheta[/tex]))
     
  15. May 10, 2010 #14

    gabbagabbahey

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    Re: Proof

    Then you would just expand (a+ib)^2 again, and compare the real and imaginary parts on both sides of the equation...
     
  16. May 10, 2010 #15
    Re: Proof

    No thanks. That's nasty.
    I'll leave the answer in rectangular form.
     
    Last edited: May 10, 2010
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