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Find a complex number z which satisfies a particular condition

  1. Nov 18, 2013 #1
    Hello guys!

    1. The problem statement, all variables and given/known data
    The question is like this:

    If ##z=\frac{a}{b}## and ##\frac{1}{a+b}=\frac{1}{a}+\frac{1}{b}##, find ##z##.

    3. The attempt at a solution
    This question is challenging for me because I don't know exactly where to start. The latter condition stated, the sum of the reciprocal of two number is equal to the sum of their reciprocal is already intuitively false for me. You just don't do that kind of operation everyday, and I think it's false for real number. I tried several complex numbers by putting ##a## and ##b## conjugate to each other and see if the condition is true. It's just not true.

    I try expressing ##a## and ##b## in rectangular form.

    If for example ##a=x+iy##, and ##b=r+is##. (1)
    $$\frac{1}{x+iy+r+is}=\frac{(x+r)-i(y+s)}{(x+r)^2+(y+s)^2}$$

    On the other hand.
    $$\frac{1}{x+iy}=\frac{x-iy}{x^2+y^2}$$
    $$\frac{1}{r+is}=\frac{r-is}{r^2+s^2}$$

    To sum them up: (2)
    $$\frac{(r^2+s^2)(x-iy)+(x^2+y^2)(r-is)}{(x^2+y^2)(r^2+s^2)}$$

    I don't see how (1) and (2) can be equal somehow by cleverly choosing the right ##a## and ##b##. On the denominator of (1) you already get addition, but then in (2) you have multiplication. Maybe if ##(x^2+y^2)## and ##(r^2+s^2)## are both ##2##, we can have the point where addition is equal to multiplication. I am not certain and I hope you can enlighten me on this.

    Thank You
     
  2. jcsd
  3. Nov 18, 2013 #2

    tiny-tim

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    Hello Seydlitz! :smile:

    Have you been at the coffee? :wink:

    1/(a+b) = 1/a + 1/b = (a + b)/ab … and solve!​
     
  4. Nov 18, 2013 #3
    Strange, I've tried it before and it leads to ##ab=1##?
     
  5. Nov 18, 2013 #4

    tiny-tim

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    ?? :confused:

    try it again …

    you should get a/b = something :smile:
     
  6. Nov 18, 2013 #5
    Woops..I made bad mistake on the way. I cancel (a+b) blatantly.

    Is ##\frac{a}{b}=\frac{-b}{a+b}##?
     
  7. Nov 18, 2013 #6

    tiny-tim

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    nooo :redface:

    show us your calculations :smile:
     
  8. Nov 18, 2013 #7

    D H

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    Try that again, tiny-tim.

    What Seydlitz posted is correct. It's just not very useful.
     
  9. Nov 18, 2013 #8
    So the only equation is this right?
    ##\frac{1}{a+b}=\frac{a+b}{ab}##
    ##\frac{a}{a+b}=\frac{a+b}{b}##
    ##\frac{a}{a+b}=\frac{a}{b}+1##
    ##\frac{a}{a+b}-1=\frac{a}{b}##
    ##\frac{a-(a-b)}{a+b}=\frac{a}{b}##
    ##\frac{-b}{a+b}=\frac{a}{b}##

    Are we expecting a numerical or symbolic answer to this question?
    (I haven't eaten yet..)
     
  10. Nov 18, 2013 #9

    tiny-tim

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    ahh!

    Seydlitz, 1/a+b = (a+b)/ab

    => ab = (a+b)2

    carry on from there :smile:
     
  11. Nov 18, 2013 #10
    Hint: in the end substitute z=a/b to make the calculation simpler.
     
  12. Nov 18, 2013 #11

    D H

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    One more hint: a/b = ab * what? Now multiply both sides of ab=something by this what to get an expression for a/b=z.
     
  13. Nov 18, 2013 #12
    $$z=\frac{-1\pm i\sqrt{3}}{2}$$

    Huzzzah!!

    Thank you everybody, especially tiny-tim for being particularly patient with me. I'm that clunky sometimes haha.
     
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