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Complex Real Number Problem

  1. Jul 7, 2009 #1
    Suppose that a and b are real numbers, not both 0. Find real numbers c and d such that


    What do they mean find c & d? That's two unknowns is it not?

    Anyway. My attempt:

    [tex]\begin{array} \frac{1}/({a+bi})=c+di\\ \Rightarrow (a+bi)(c+di)=1\\\Rightarrow ac-bd+(ad+bc)i=1\\\Rightarrow \frac{1+bd-ac}{ad+bc}=i\end{array}[/tex]

    Now I know that I could square both sides to get rid of i, but I am not sure how whether that will help or hurt.
  2. jcsd
  3. Jul 7, 2009 #2


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    Squaring won't get rid of the i. Use complex conjugates. Multiply the left side by (a-bi)/(a-bi). Equate real and imaginary parts.
  4. Jul 7, 2009 #3
    Okay. But why won't squaring both sides of my last line get rid of i? I mean, it will leave me dead in the water, but the RHS will surely be -1 right? (Just want to be sure i am not missing something big here)
  5. Jul 7, 2009 #4
    If I multiply the original EQ by the conjugate I get


    I am not sure what you mean by equate the real and imaginary parts? Sorry, I have never really used complex numbers :redface:

    on the LHS, I have the complex number [tex]\frac{a}{a^2+b^2}+\frac{-b}{a^2+b^2}i[/tex]

    Does that form help my plight?
  6. Jul 7, 2009 #5


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    Sure. The real part of the left side is a/(a^2+b^2), the real part of the right side is c. So c=a/(a^2+b^2). (a,b,c,d are REAL). Now equate the imaginary parts of both sides. You are supposed to find c and d in terms of a and b.
  7. Jul 7, 2009 #6
    Ohhhhh. i see where my confusion lies. a and be are real numbers, BUT we refer to the quantity (b*i) as the imaginary PART right?

    i is the only imaginary number but b*i is the imaginary part
  8. Jul 7, 2009 #7


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    If you have a complex number a+ib, then a is the real part and b is the imaginary part not ib.
  9. Jul 7, 2009 #8


    Staff: Mentor

    In other words, the imaginary part is the coefficient of i. That means that both the real part and the imaginary part are real numbers.
  10. Jul 7, 2009 #9


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    and bi is an imaginary number, not only i.
  11. Jul 7, 2009 #10


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    Unfortunately, you can find as many authoritative statements that the "imaginary part" of a+ bi is "bi" as you can that it is "b". Fortunately, it is not a critical distinction.
    [tex]\frac{a- bi}{a^2+ b^2}= \frac{a}{a^2+ b^2}- \frac{bi}{a^2+ b^2}= c+ di[/tex]
    [tex]\frac{a}{a^2+ b^2}= c[/tex]
    [tex]\frac{-b}{a^2+ b^2}= d[/tex]
    no matter what you call them!
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